poj 3177 Redundant Paths(Tarjan，边双连通分量)

来源：互联网发布：网络盒子加密软件编辑：程序博客网时间：2024/06/05 08:23

kuangbin模板中的例题，我把模板里的InStack去掉了。
思路：
这里写图片描述

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int MAXN = 5010;const int MAXM = 20010;struct Edge{    int to,next;    bool cut;} edge[MAXM];int head[MAXN],tot;int Low[MAXN],Dfn[MAXN],Stack[MAXN],Belong[MAXN];int Index,top;int block;int bridge;void addedge(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    edge[tot].cut = false;    head[u] = tot++;}void init(){    memset(head,-1,sizeof(head));    tot = 0;}void Tarjan(int u, int pre){    int v;    Low[u] = Dfn[u] = ++Index;    Stack[top++] = u;    for(int i = head[u]; i != -1; i = edge[i].next)    {        v = edge[i].to;        if(v == pre) continue;        if(!Dfn[v])        {            Tarjan(v,u);            if(Low[u] > Low[v]) Low[u] = Low[v];            if(Low[v] > Dfn[u])            {                bridge++;                edge[i].cut = true;                edge[i^1].cut = true;            }        }        else if(Low[u] > Dfn[v])            Low[u] = Dfn[v];    }    if(Low[u] == Dfn[u])    {        block++;        do        {            v = Stack[--top];            Belong[v] = block;        }while(v != u);    }}int du[MAXN];void solve(int n){    memset(Dfn,0,sizeof(Dfn));    Index = top = block = 0;    Tarjan(1,0);    int ans = 0;    memset(du,0,sizeof(du));    for(int i = 1; i <= n; ++i)    {        for(int j = head[i]; j != -1; j = edge[j].next)        {            if(edge[j].cut)                du[Belong[i]]++;        }    }    for(int i = 1; i <= block; ++i)        if(du[i] == 1)            ++ans;    printf("%d\n",(ans+1)/2);}int main(){    int n,m,u,v;    while(scanf("%d %d",&n,&m) != EOF)    {        init();        while(m--)        {            scanf("%d %d",&u,&v);            addedge(u,v);            addedge(v,u);        }        solve(n);    }    return 0;}

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