POJ 3177 Redundant Paths（求桥）

题目链接

题意

给你一个联通图，求最小加几条边，使图变成双连通图

思路

求桥，将双连通分量缩点，以桥为边构成新图，图中度数为1的节点都需要加边，所以答案为度数为1节点除以2向上取整。

总结

无多重边，只需记录父节点，判断是否走回去即可
多重边求桥链式前向星建图，通过同一条边id差1避免同一条边重复遍历

代码

#include <stdio.h>
#include <string.h>
#include <vector>
#include <stack>
using namespace std;

vector<int> e[5005];
stack<int> sta;
int d[5005], dfn[5005], low[5005], tim, col[5005], x[5005][2], cnt, bri;

void tarjan(int u, int fa)
{
    sta.push(u);
    dfn[u] = low[u] = ++tim;
    for(int i = 0; i < e[u].size(); ++i)
    {
        int v = e[u][i];
        if(!dfn[v])
        {
            tarjan(v,u);
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u]) x[bri][0] = u, x[bri++][1] = v;
        }
        else if(v != fa) low[u] = min(low[u], dfn[v]); // 判断防止回头，真滴巧妙
    }
    if(dfn[u] == low[u])
    {
        ++cnt;
        while(1)
        {
            int now = sta.top();
            sta.pop();
            col[now] = cnt;
            if(now == u) break;
        }
    }
}

int main()
{
    int n, m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i = 1; i <= n; ++i) e[i].clear();
        while(m--)
        {
            int u, v;
            scanf("%d%d",&u,&v);
            e[v].push_back(u);
            e[u].push_back(v);
        }
        for(int i = 1; i <= n; ++i) d[i] = dfn[i] = low[i] = 0;
        while(!sta.empty()) sta.pop();
        tim = 0;
        cnt = 0;
        bri = 0;
        tarjan(1,-1);
        int ans = 0;
        for(int i = 0; i < bri; ++i) ++d[col[x[i][0]]], ++d[col[x[i][1]]];
        for(int i = 1; i <= cnt; ++i) if(d[i] == 1) ++ans;
        printf("%d\n",(ans+1)/2);
    }
    return 0;
}