HDU 2767Proving Equivalences 强连通分量

来源：互联网发布：2017网络视听大会编辑：程序博客网时间：2024/05/16 08:24

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2767

题意：题目描述很繁杂，大概意思就是求最少加几条边可以使图只有一个强连通分量。

思路：用tarjan算法求强连通分量缩点，统计每个点的入度和出度，最后输出入度为0和出度为0的点的个数中的较大值，至于为什么这样做，没有证明，不过可以自己实验一下

总结：跟poj某道题差不多

#include<iostream>#include<algorithm>#include<cstdio>#include<queue>#include<map>#include<vector>#include<cstring>#include<cctype>#include<cmath>using namespace std;typedef long long ll;const int N = 20100;struct edge{    int to, next;} G[N*5];int dfn[N], low[N], scc[N], st[N], head[N];int index, cnt, top, num;bool vis[N];int n, m;void init(){    memset(head, -1, sizeof head);    memset(dfn, -1, sizeof dfn);    memset(vis, 0, sizeof vis);    index = cnt = top = num = 0;}void add_edge(int v, int u){    G[cnt].to = u;    G[cnt].next = head[v];    head[v] = cnt++;}void tarjan(int v){    dfn[v] = low[v] = index++;    vis[v] = true;    st[top++] = v;    int u;    for(int i = head[v]; i != -1; i = G[i].next)    {        u = G[i].to;        if(dfn[u] == -1)        {            tarjan(u);            low[v] = min(low[v], low[u]);        }        else if(vis[u])            low[v] = min(low[v], dfn[u]);    }    if(dfn[v] == low[v])    {        num++;        do        {            u = st[--top];            vis[u] = false;            scc[u] = num;        }        while(u != v);    }}void slove(){    for(int i = 1; i <= n; i++)        if(dfn[i] == -1)            tarjan(i);    if(num == 1)    {        printf("0\n");        return;    }    int outdeg[N], indeg[N];    memset(outdeg, 0, sizeof outdeg);    memset(indeg, 0, sizeof indeg);    for(int i = 1; i <= n; i++)        for(int j = head[i]; j != -1; j = G[j].next)            if(scc[i] != scc[G[j].to])                outdeg[scc[i]]++, indeg[scc[G[j].to]]++;    int in0 = 0, out0 = 0;    for(int i = 1; i <= num; i++)    {        if(outdeg[i] == 0) out0++;        if(indeg[i] == 0) in0++;    }    printf("%d\n", max(in0, out0));}int main(){    int t, a, b;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        init();        for(int i = 0; i < m; i++)        {            scanf("%d%d", &a, &b);            add_edge(a, b);        }        slove();    }    return 0;}

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