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  • 1827 tarjan+缩点

    #include<stdio.h>
    #include<stack>
    #include<iostream>
    #include<string.h>
    #include<vector>
    #include<string>
    using namespace std;
    #define inf 1000000000
    #define N 2100
    vector<int>h[N];//
    stack<int>q;///
    int dfn[N],low[N],index,indegree[N],a[N],visit[N],suo[N],fee[N],yong;
    int min(int a,int b) {
    return a>b?b:a;
    }
    void tarjan(int u) {
    int i;
    dfn[u]=low[u]=++index;
    visit[u]=1;
    q.push(u);
    for(i=0;i<h[u].size();i++) {
    int v=h[u][i];
    if(!dfn[v]) {
    tarjan(v);
    low[u]=min(low[u],low[v]);//
    }
    else
    if(visit[v]==1) //在寨中
    low[u]=min(low[u],low[v]);
    }
    if(dfn[u]==low[u]) {//如果存在一个强连通分量就将里面的出寨
    yong++;
    int t;
    do {
       t=q.top();
    q.pop();
    visit[t]=2;//已经浏览过了
    suo[t]=yong;//讲一个连通分量置为一个点
                if(fee[yong]>a[t])//求出缩点中最小的费用
    fee[yong]=a[t];
    }while(t!=u);

    }
    }
    int main() {
    int n,m,i,j,count,num;
    while(scanf("%d%d",&n,&m)!=EOF) {
    while(!q.empty()) 
    q.pop();
    for(i=1;i<=n;i++) {//输入和初始化
    scanf("%d",&a[i]);
    h[i].clear();
    fee[i]=inf;
    }
    while(m--) {
    scanf("%d%d",&i,&j);//
    h[i].push_back(j);//

    memset(visit,0,sizeof(visit));
    index=0;yong=0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    for(i=1;i<=n;i++)
    if(visit[i]!=2)//是否已经入过寨
    tarjan(i);
    count=0;
    memset(indegree,0,sizeof(indegree));
    for(i=1;i<=n;i++) 
    for(j=0;j<h[i].size();j++) 
      if(suo[i]!=suo[h[i][j]])//如果不是同一个缩点
      indegree[suo[h[i][j]]]++;
      num=0;
      for(i=1;i<=yong;i++)
      if(indegree[i]==0) {//找入度为零的点
      count+=fee[i];
      num++;
      }
      printf("%d %d ",num,count);
    }
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/thefirstfeeling/p/4410860.html
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