【Codeforces 1027D】Mouse Hunt

【链接】 我是链接,点我呀:)

【题意】

题意

【题解】

先求出来强连通分量。
每个联通分量里面,显然在联通块的尽头(没有出度)放一个捕鼠夹就ok了

【代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define ri(x) scanf("%d",&x)

#define rl(x) scanf("%lld",&x)

#define rs(x) scanf("%s",x)

#define oi(x) printf("%d",x)

#define ol(x) printf("%lld",x)

#define oc putchar(' ')

#define os(x) printf(x)

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

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 2e5;//节点个数

vector <int> G[N+10],g[N+10];

int n,m,tot = 0,top = 0,dfn[N+10],low[N+10],z[N+10],totn,in[N+10];

int bh[N+10];

int mi[N+10];

bool bo[N+10];

int c[N+10];

int f[N+10],f_ans[N+10];

void dfs(int x){

    dfn[x] = low[x] = ++ tot;

    z[++top] = x;

    in[x] = 1;

    int len = G[x].size();

    rep1(i,0,len-1){

        int y = G[x][i];

        if (!dfn[y]){

            dfs(y);

            low[x] = min(low[x],low[y]);

        }else

        if (in[y] && dfn[y]<low[x]){

            low[x] = dfn[y];

        }

    }

    if (low[x]==dfn[x]){

        int v = 0;

        totn++;

        mi[totn] = c[z[top]];

        while (v!=x){

            v = z[top];

            mi[totn] = min(mi[totn],c[v]);

            in[v] = 0;

            bh[v] = totn;

            top--;

        }

    }

}

int dfs1(int x){

    if (bo[x]==true) return 0;

    bo[x] = true;

    int len = g[x].size();

    if (len==0) return mi[x];

    return dfs1(g[x][0]);

}

int main(){

    #ifdef LOCAL_DEFINE

        freopen("rush_in.txt", "r", stdin);

    #endif

    ms(dfn,0);

    ms(in,0);

    tot = 0,totn = 0;

    ri(n);

    rep1(i,1,n) G[i].clear(),g[i].clear();

    rep1(i,1,n){

        cin >> c[i];

    }

    rep1(i,1,n){

        int x,y;

        ri(y);

        x = i;

        G[x].pb(y);

    }

    rep1(i,1,n)

        if (dfn[i]==0)

            dfs(i);

    rep1(i,1,n){

        int len = G[i].size();

        int xx = bh[i];

        rep1(j,0,len-1){

            int y = G[i][j];

            int yy = bh[y];

            if (xx!=yy){

                g[xx].pb(yy);

            }

        }

    }

    n = totn;

    int ans = 0;

    rep1(i,1,n) ans += dfs1(i);

    cout<<ans<<endl;

    return 0;

}