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今日SGU 5.26

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

struct Point{

    db x,y;

}ans[maxn],p1,p2,o;

Point get_mid(Point p1,Point p2){

    Point ret;

    ret.x = (p1.x + p2.x) / ;

    ret.y = (p1.y + p2.y) / ;

    return ret;

}

Point get_o(Point p1,Point p2,int n1,int n2,int n){//正多边形中点,外接圆圆心

    Point mid = get_mid(p1,p2),ret;

    db ang = PI / 2.0 - 1.0 * (n2 - n1) * PI / n;

//    向量的思想

    ret.x = mid.x - (p1.y - mid.y) * tan(ang);

    ret.y = mid.y + (p1.x - mid.x) * tan(ang);

    return ret;

}

db f(db x){//会出现-0的情况

    return fabs(x)<eps?:x;

}

int main(){

    int n,n1,n2;

    scanf("%d%d%d",&n,&n1,&n2);

    scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y);

    if(n1 > n2) swap(n1,n2), swap(p1,p2);

    n1--; n2--;

    o = get_o(p1,p2,n1,n2,n);

    db ang = 2.0 * PI / n;

    ans[n1] = p1;

//    dd(o.x)de(o.y)

    for(int i = n1+; ;++i){

        if(i%n==n1) break;

//        x0,y0绕rx0,ry0逆时针旋转a度

//        x0= (x - rx0)*cos(a) - (y - ry0)*sin(a) + rx0 ;

//        y0= (x - rx0)*sin(a) + (y - ry0)*cos(a) + ry0 ;

        ans[i%n].x = o.x + (ans[(i-+n)%n].x - o.x)*cos(-ang) - (ans[(i-+n)%n].y - o.y)*sin(-ang);

        ans[i%n].y = o.y + (ans[(i-+n)%n].x - o.x)*sin(-ang) + (ans[(i-+n)%n].y - o.y)*cos(-ang);

    }

    rep(i,,n) printf("%.8f %.8f\n",f(ans[i].x),f(ans[i].y));

    return ;

}

SGU 120

题意：给你正n边形上的两个点，让你求出正n边形上的所有点，1 - n是顺时针

收获:求正n边形中心或者说是外接圆的圆心，向量旋转的公式[x*cosA-y*sinA  x*sinA+y*cosA]（向量逆时针旋转A）

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

struct Point{

    db x,y;

}ans[maxn],p1,p2,o;

Point get_mid(Point p1,Point p2){

    Point ret;

    ret.x = (p1.x + p2.x) / ;

    ret.y = (p1.y + p2.y) / ;

    return ret;

}

Point get_o(Point p1,Point p2,int n1,int n2,int n){//正多边形中点,外接圆圆心

    Point mid = get_mid(p1,p2),ret;

    db ang = PI / 2.0 - 1.0 * (n2 - n1) * PI / n;

//    向量的思想

    ret.x = mid.x - (p1.y - mid.y) * tan(ang);

    ret.y = mid.y + (p1.x - mid.x) * tan(ang);

    return ret;

}

db f(db x){//会出现-0的情况

    return fabs(x)<eps?:x;

}

int main(){

    int n,n1,n2;

    scanf("%d%d%d",&n,&n1,&n2);

    scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y);

    if(n1 > n2) swap(n1,n2), swap(p1,p2);

    n1--; n2--;

    o = get_o(p1,p2,n1,n2,n);

    db ang = 2.0 * PI / n;

    ans[n1] = p1;

//    dd(o.x)de(o.y)

    for(int i = n1+; ;++i){

        if(i%n==n1) break;

//        x0,y0绕rx0,ry0逆时针旋转a度

//        x0= (x - rx0)*cos(a) - (y - ry0)*sin(a) + rx0 ;

//        y0= (x - rx0)*sin(a) + (y - ry0)*cos(a) + ry0 ;

        ans[i%n].x = o.x + (ans[(i-+n)%n].x - o.x)*cos(-ang) - (ans[(i-+n)%n].y - o.y)*sin(-ang);

        ans[i%n].y = o.y + (ans[(i-+n)%n].x - o.x)*sin(-ang) + (ans[(i-+n)%n].y - o.y)*cos(-ang);

    }

    rep(i,,n) printf("%.8f %.8f\n",f(ans[i].x),f(ans[i].y));

    return ;

}

SGU 186

题意：给你n个链子，每个链子有li个串，你可以在一个时间段里，把一个串从一个链子拆下来，然后用它把起来的链子连起来，

问你最少用多少时间来把所有的链子拼成一个链子

收获：贪心

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int l[maxn];

int main(){

    int n;

    scanf("%d",&n);

    rep(i,,n) scanf("%d",l+i);

    sort(l,l+n);

    int sum = n - ,ans = ;;

    rep(i,,n) {

        while(l[i]--&&sum){

            ans++;

            sum--;

            if(!sum) break;

        }

        if(!sum) break;

        sum--;

    }

    printf("%d\n",ans);

    return ;

}

SGU 226

题意：给你n个点，m个边，每个边有一个颜色，他要你找到点1到点n的最短距离，路径上的相邻两条边的颜色不能一样

收获：spfa + 颜色限制

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,m;

struct edge{

    int v,nt,c;

}e[maxn*maxn];

int pre[maxn],tot;

int dis[][maxn];

bool in[maxn];

void init(){

    tot = ;

    mt(in,false);mt(pre,-);

    rep(i,,) rep(j,,n+) dis[i][j] = inf;

    dis[][] = dis[][] = dis[][] = dis[][] = ;

}

void add(int u,int v,int c){

    e[tot].v = v;

    e[tot].c = c;

    e[tot].nt = pre[u];

    pre[u] = tot++;

}

void spfa(){

    queue<int> q;

    q.push();

    while(sz(q)){

        int u = q.front(); q.pop(); in[u] = false;

        de(u)

        for(int i = pre[u]; ~i; i = e[i].nt){

            int v = e[i].v , col = e[i].c;

            rep(j,,) if(j != col) {

                if(dis[col][v] > dis[j][u] + ){

                    dis[col][v] = dis[j][u] + ;

                    if(!in[v]) q.push(v),in[v] = true;

                }

            }

        }

    }

}

int main(){

    scanf("%d%d",&n,&m);

    init();

    rep(i,,m) {

        int u,v,c;

        scanf("%d%d%d",&u,&v,&c);

        add(u,v,c);

    }

    spfa();

    int mn = inf;

//    rep(j,1,n+1) rep(i,1,4) printf("%d%c",dis[i][j]," \n"[i==3]);

    rep(i,,) mn = min(dis[i][n],mn);

    printf("%d\n",mn==inf?-:mn);

    return ;

}

/*

    5 5

    1 2 2

    1 3 2

    2 3 3

    2 4 1

    3 5 2

    3 3

    1 2 2

    2 2 1

    2 3 2

*/

SGU 174

题意：给你n个线段，线段只会在端点相交，然后问你到第几步的时候 ，线段围成一个封闭区间

收获：并查集无向图判环

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e5+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n;

int f[maxn<<];

void init(){

    rep(i,,*n+) f[i] = i;

}

map<pii,int> m;

int id = ;

int get_id(pii x){

    if(m.count(x)) return m[x];

    m[x] = id; ++id;

    return m[x];

}

int fd(int x) { return x==f[x]?x:f[x]=fd(f[x]); }

int main(){

    int ans = ;

    scanf("%d",&n);

    init();

    rep(i,,n+){

        int x1,y1,x2,y2;

        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

        int id1 = get_id(mp(x1,y1));

        int id2 = get_id(mp(x2,y2));

        int f1 = fd(id1), f2 = fd(id2);

//        dd(id1)dd(f1)dd(id2)de(f2)

        if(f1 == f2) {

            if(!ans) ans = i;

        }

        else f[f1] = f2;

    }

    printf("%d\n",ans);

    return ;

}

SGU 224

题意：n*n的矩形放k个皇后

收获：位运算实现n皇后

大佬的题解：http://www.matrix67.com/blog/archives/266

我曾经傻傻的超时代码

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=; if(b<) return ; for(;b;b>>=){if(b&)res=res*a;a=a*a;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,k;

int ans = ;

int dp[];

int fd_pos(int now){

    if(!now) return -;

    rep(i,,n) if(kpow(,i) == now) return i;

}

bool ok(int x,int nowpos){

    rep(i,,x){

        int pos = fd_pos(dp[i]);

        if(pos == -) continue;

        if(pos == nowpos) return false;

        if(abs(x - i) == abs(nowpos - pos)) return false;

    }

    return true;

}

void dfs(int x,int d){

    if(x == n+){

        if(!d) ans++;

        return;

    }

    rep(i,,n){

        int now = kpow(,i);

        if(!ok(x,i) || !d) continue;

        dp[x] = now;

        if(d) dfs(x + ,d - );

    }

    dp[x] = ;

    dfs(x + ,d);

}

int main(){

    scanf("%d%d",&n,&k);

    if(!k) return puts(""),;

    if(k > n) return puts(""),;

    dfs(,k);

    printf("%d\n",ans);

    return ;

}

AC代码

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=; if(b<) return ; for(;b;b>>=){if(b&)res=res*a;a=a*a;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,k;

int ans = ;

int lowbit(int x) { return x & (-x); }

void dfs(int x,int l,int r,int d,int num){

    if(num == k){

        ans++;

        return;

    }

    if(x == n + ) return;

    int canpos = (( << n)- ) & (~(r | l | d));

    while(canpos) {

        int t = lowbit(canpos);

        canpos -= t;

        dfs(x + , (l | t) << , (r | t) >> , d | t,num + );

    }

    dfs(x + , l << , r >> ,d,num);

}

int main(){

    scanf("%d%d",&n,&k);

    if(!k) return puts(""),;

    if(k > n) return puts(""),;

    dfs(,,,,);

    printf("%d\n",ans);

    return ;

}
05-11 17:10
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