hdu 3849 By Recognizing These Guys (割边/桥)

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题目链接: hdu 3849

题目大意: 给出无向图，找出所有的割边

并且按边输入的顺序输出

如果图不连通则直接输出0

解题思路: 先用map或者Hash把字符串转化为编号

根据编号建立无向图

Tarjan搜索无向图的桥

判断Tarjan的深度，小于n说明是不连通图

代码:

//Final   Tarjan搜索桥#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#include <map>#include <string>using namespace std;#define MAX 10005#define MIN(a,b) a<b?a:bstruct snode{int to,next,num;}edge[MAX*20];char ZM[MAX*10][16];char ZM2[MAX*10][16];char ch1[MAX],ch2[MAX];int visit[MAX],low[MAX],dnf[MAX],pre[MAX],ans[MAX*10],Index,high,kk;void Add_edge(int a,int b,int m){edge[Index].num=m;edge[Index].to=b;edge[Index].next=pre[a];pre[a]=Index++;return ;}void Tarjan(int u,int father){int v,vv;for(v=pre[u];v!=-1;v=edge[v].next){vv=edge[v].to;if(!visit[vv]){visit[vv]=1;low[vv]=dnf[vv]=++high;            Tarjan(vv,u);            low[u]=MIN(low[u],low[vv]);            if(low[vv]>dnf[u])               //low[vv]>dnf[u]就是桥            { ans[kk++]=edge[v].num;            }}        else if(vv!=father)                  //**        {            low[u]=MIN(low[u],dnf[vv]);                    }}}int main(){    int n,m,i,k,t,a,b;scanf("%d",&t);while(t--){        map<string,int> HashZS;Index=kk=0;memset(pre,-1,sizeof(pre));memset(dnf,0,sizeof(dnf));memset(visit,0,sizeof(visit));scanf("%d%d",&n,&m);for(i=0,k=0;i<m;i++){scanf("%s%s",ch1,ch2);if(HashZS[ch1]==0)       //map把字符串转换成编号{HashZS[ch1]=++k;}if(HashZS[ch2]==0)       //map把字符串转换成编号{HashZS[ch2]=++k;}strcpy(ZM[i],ch1);strcpy(ZM2[i],ch2);a=HashZS[ch1];b=HashZS[ch2];Add_edge(a,b,i);Add_edge(b,a,i);}low[1]=dnf[1]=visit[1]=high=1;Tarjan(1,-1);if(high!=n)                  //如果Tarjan的深度小于n则说明图不连通{printf("0\n");continue;}        sort(ans,ans+kk);            //根据边输入的顺序排序printf("%d\n",kk);for(i=0;i<kk;i++){printf("%s %s\n",ZM[ans[i]],ZM2[ans[i]]);}}return 0;}