NOIP2014

DAY1

生活大爆炸版石头剪刀布

　　直接模拟即可。

// codevs3716

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<cmath>

#define LL long long

#define MOD 1000000007

#define inf 2147483640

#define Pi acos(-1.0)

#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);

using namespace std;

const int maxn=2000;

int c[5][5]={{0,0,1,1,0},

			 {1,0,0,1,0},

			 {0,1,0,0,1},

			 {0,0,1,0,1},

			 {1,1,0,0,0}};

int a[maxn],b[maxn];

int main() {

	int n,na,nb;

	scanf("%d%d%d",&n,&na,&nb);

	for (int i=0;i<na;i++) scanf("%d",&a[i]);

	for (int i=0;i<nb;i++) scanf("%d",&b[i]);

	int A=0,B=0;

	for (int i=0;i<n;i++) {

		A+=c[a[i%na]][b[i%nb]];

		B+=c[b[i%nb]][a[i%na]];

	}

	printf("%d %d",A,B);

	return 0;

}

联合权值

　　一开始无脑枚举，因为只需要dfs1层，感觉完全不虚，结果被菊花树卡得只有70分，于是怒水一发树形dp。

// codevs3728

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<cmath>

#define LL long long

#define MOD 10007

#define inf 2147483640

#define Pi acos(-1.0)

#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);

using namespace std;

const int maxn=200010;

struct edge {int to,next;}e[maxn<<1];

int head[maxn],n,cnt,w[maxn],ans1,ans2;;

void link(int u,int v) {

	e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;

	e[++cnt].to=u;e[cnt].next=head[v];head[v]=cnt;

}

void dfs(int x,int fa,int f) {

	ans1=(ans1+w[x]*w[f]);ans2=max(ans2,w[x]*w[f]);

	int x1=0,x2=0;

	for (int i=head[x];i;i=e[i].next) if (e[i].to!=fa) {

			dfs(e[i].to,x,fa);

			ans2=max(ans2,x2*w[e[i].to]);

			ans1=(ans1+x1*w[e[i].to])%MOD;

			x1=(x1+w[e[i].to])%MOD;x2=max(x2,w[e[i].to]);

		}

}

int main() {

	scanf("%d",&n);

	for (int u,v,i=1;i<n;i++) {

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

		link(u,v);

	}

	for (int i=1;i<=n;i++) scanf("%d",&w[i]);

	dfs(1,0,0);

	printf("%d %d",ans2,ans1*2%MOD);

	return 0;

}

飞扬的小鸟

　　一开始打了个nm²加队列乱搞可以获得85分的高分哦！正解背包。

// codevs3729

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<cmath>

#define LL long long

#define inf 100000000

#define Pi acos(-1.0)

#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);

using namespace std;

const int maxn=10010,maxm=1010;

struct data {int p,l,r;}t[maxn];

int n,m,K,u[maxn],d[maxn],l[maxn],r[maxn],f[maxn][maxm];

int main() {

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

	for (int i=0;i<n;i++) scanf("%d%d",&u[i],&d[i]);

	for (int i=0;i<=n;i++) l[i]=0,r[i]=m+1;

	for (int x,i=1;i<=K;i++) {

		scanf("%d",&x);

		scanf("%d%d",&l[x],&r[x]);

	}

	int cnt=0;

	for (int i=1;i<=n;i++) {

		for (int j=1;j<=m;j++) {

			f[i][j]=inf;

			if (j>u[i-1]) f[i][j]=min(f[i][j],min(f[i-1][j-u[i-1]],f[i][j-u[i-1]])+1);

		}

		for (int j=m-u[i-1];j<=m;j++) f[i][m]=min(f[i][m],min(f[i-1][j],f[i][j])+1);

		for (int j=l[i]+1;j<=r[i]-1;j++)

			if (j+d[i-1]<=m) f[i][j]=min(f[i][j],f[i-1][j+d[i-1]]);

		for (int j=1;j<=l[i];j++) f[i][j]=inf;

		for (int j=r[i];j<=m;j++) f[i][j]=inf;

		int flag=0;

		for (int j=1;j<=m;j++) if (f[i][j]<inf) {flag=1;break;}

		if (!flag) {printf("0\n%d",cnt);return 0;}

		else if (r[i]!=m+1) cnt++;

	}

	int ans=inf;

	for (int i=1;i<=m;i++) ans=min(ans,f[n][i]);

	printf("1\n%d",ans);

	return 0;

}

DAY2

无线网络发射选址

　　无脑枚举。

// codevs3730

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<queue>

#define LL long long

#define MOD 10007

#define inf 2147483640

#define Pi acos(-1.0)

#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);

using namespace std;

int f[200][200],n,d;

int main() {

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

	for (int x,y,k,i=1;i<=n;i++) {

		scanf("%d%d%d",&x,&y,&k);

		f[x][y]=k;

	}

	int ans=0,tot=0;

	for (int i=0;i<=128;i++)

		for (int j=0;j<=128;j++) {

			int cnt=0;

			for (int k=max(i-d,0);k<=min(i+d,128);k++)

				for (int l=max(j-d,0);l<=min(j+d,128);l++) cnt+=f[k][l];

			if (ans==cnt) tot++;

			else if (ans<cnt) ans=cnt,tot=1;

		}

	printf("%d %d",tot,ans);

	return 0;

}

寻找道路

　　写得奇丑无比。。还Wa了两发，数组开小了→_→。。先反向连边处理哪些点能走哪些点不能走，然后Dijkstra。

// codevs3731

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<queue>

#define LL long long

#define MOD 10007

#define inf 2147483640

#define Pi acos(-1.0)

#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);

using namespace std;

const int maxn=10010,maxm=200010;

struct edge {int to,next;}e[maxm<<1];

struct data {

	int num,w;

	friend bool operator < (const data a,const data b) {

		return a.w>b.w;

	}

};

int head[maxn],f[maxn],vis[maxn],dis[maxn],u[maxm],v[maxm];

int n,m,cnt,s,t;

void link(int u,int v) {

	e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;

}

void dfs(int x) {

	vis[x]=1;

	for (int i=head[x];i;i=e[i].next) if (!vis[e[i].to]) dfs(e[i].to);

}

void Dijkstra() {

	priority_queue<data> q;

	data x=(data){s,0},y;

	for (int i=1;i<=n;i++) dis[i]=inf,vis[i]=0;

	dis[s]=0;q.push(x);

	while (!q.empty() && !vis[t]) {

		x=q.top();q.pop();

		if (vis[x.num]) continue;

		vis[x.num]=1;

		for (int i=head[x.num];i;i=e[i].next)

			if (f[e[i].to] && dis[e[i].to]>x.w+1) {

				dis[e[i].to]=y.w=x.w+1;

				y.num=e[i].to;

				q.push(y);

			}

	}

}

int main() {

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

	for (int i=1;i<=m;i++) {

		scanf("%d%d",&u[i],&v[i]);

		link(v[i],u[i]);

		//link(u,v);

	}

	scanf("%d%d",&s,&t);swap(s,t);

	dfs(s);

	memset(head,0,sizeof(head));cnt=0;

	for (int i=1;i<=m;i++) link(u[i],v[i]);

	for (int i=1;i<=n;i++) {

		f[i]=vis[i];

		for (int j=head[i];j;j=e[j].next) f[i]&=vis[e[j].to];

	}

	swap(s,t);

	Dijkstra();

	printf("%d",dis[t]==inf ? -1 : dis[t]);

	return 0;

}

解方程

　　一开始一直纠结怎么优化高精度，一直无果。。模了题解没想到是这这样的结果→_→。

　　我们发现若将等式左侧模上一个数等于0，那么有可能这个x是解。而在模M的意义下,f[x]与f[x+M]的值是一样的。于是我们就随便搞5个素数，分别预处理出从1~M-1的范围中的解，然后枚举x判断即可。

// codevs3732

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<queue>

#define LL long long

#define MOD 10007

#define inf 2147483640

#define Pi acos(-1.0)

#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);

using namespace std;

const int maxn=1000010;

int M[5]={9973,9931,9941,9949,9967};

int ans[maxn],a[5][maxn],res[5][maxn],pre[5][maxn];

int n,m;

int cal(int t,int x) {

	int sum=0;

	for (int i=0;i<=n;i++) sum=(sum+a[t][i]*pre[t][i])%M[t];

	if (sum<0) sum+=M[t];

	return sum;

}

bool check(int x) {

	for (int t=0;t<5;t++) if (res[t][x%M[t]]!=0) return 0;

	return 1;

}

int main() {

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

	char ch[10010];

	for (int i=0;i<=n;i++) {

		scanf("%s",ch+1);

		int l=strlen(ch+1);

		bool flag=0;

		for (int t=0;t<5;t++) {

			if (ch[1]!='-') a[t][i]=ch[1]-'0';

			else a[t][i]=0,flag=1;

		}

		for (int t=0;t<5;t++) {

			for (int k=2;k<=l;k++) a[t][i]=(a[t][i]*10+ch[k]-'0')%M[t];

			if (flag) a[t][i]=-a[t][i];

		}

	}

	for (int t=0;t<5;t++)

		for (int x=1;x<M[t];x++) {

			pre[t][0]=1;

			for (int i=1;i<=n;i++) pre[t][i]=(pre[t][i-1]*x)%M[t];

			res[t][x]=cal(t,x);

		}

	for (int i=1;i<=m;i++) if (check(i)) ans[++ans[0]]=i;

	printf("%d\n",ans[0]);

	for (int i=1;i<=ans[0];i++) printf("%d\n",ans[i]);

	return 0;

}