POJ-3177(tarjan缩点）

题目大意：至少加多少条边使得每两个点都有至少两种走法可以走
题解思路：利用tarjan缩点后找有几点只有一条边跟这个点跟它连接然后用最少的边把这些点都变成有两条边就ok.
如果不缩点那么以下样例就不用加边了
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
注意记得判断重边

题目链接

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<stack>using namespace std;const int mx = 1e4+5;struct node{    int next;    int v;}E[mx<<3];stack<int>s;int n,m;int pre[mx],low[mx],head[mx];int deg[mx],sccno[mx];int dfn,tot,bri,cnt;void init(){    memset(pre,0,sizeof(pre));    memset(head,0,sizeof(head));    memset(deg,0,sizeof(deg));    dfn = tot = bri = cnt = 0;}void add(int u,int v){    E[++tot].v = v;    E[tot].next = head[u];    head[u] = tot;}void dfs(int u,int fa){    pre[u] = low[u] = ++dfn;    s.push(u);    int flag = 0;    for(int i = head[u]; i ; i = E[i].next){        int v = E[i].v;        if(!pre[v]){            dfs(v,u);            low[u] = min(low[v],low[u]);        }        else if(v != fa || flag)            low[u] = min(low[v],low[u]);        if(v == fa)            flag = 1;    }    if(low[u] == pre[u]){        cnt++;        while(1){            int v = s.top();            s.pop();            sccno[v] = cnt;            if(v == u)  break;        }    }}void solve(){    dfs(1,-1);    for(int u = 1; u <= n; u++){        for(int i = head[u]; i ; i = E[i].next){            int v = E[i].v;            if(sccno[v] != sccno[u]){                deg[sccno[v]]++;            }        }    }    for(int i = 1; i <= cnt; i++)        if(deg[i] == 1)            bri++;    printf("%d\n",(bri+1)/2);}int main(){    while(~scanf("%d%d",&n,&m)){        init();        for(int i = 1; i <= m; i++){            int u,v;            scanf("%d%d",&u,&v);            add(u,v);            add(v,u);        }        solve();    }    return 0;}