Caocao‘s Bridges

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8476 Accepted Submission(s): 2604

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4738

Description:

Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn‘t give up. Caocao‘s army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river, and based on those islands, Caocao‘s army could easily attack Zhou Yu‘s troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao‘s army could be deployed very conveniently among those islands. Zhou Yu couldn‘t stand with that, so he wanted to destroy some Caocao‘s bridges so one or more islands would be seperated from other islands. But Zhou Yu had only one bomb which was left by Zhuge Liang, so he could only destroy one bridge. Zhou Yu must send someone carrying the bomb to destroy the bridge. There might be guards on bridges. The soldier number of the bombing team couldn‘t be less than the guard number of a bridge, or the mission would fail. Please figure out as least how many soldiers Zhou Yu have to sent to complete the island seperating mission.

Input:

There are no more than 12 test cases.

In each test case:

The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N² )

Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )

Output:

For each test case, print the minimum soldier number Zhou Yu had to send to complete the mission. If Zhou Yu couldn‘t succeed any way, print -1 instead.

Sample Input:

3 3
1 2 7
2 3 4
3 1 4
3 2
1 2 7
2 3 4
0 0

Sample Output:

-1
4

題意：

給出一個無向圖，然後每條邊都有邊權，求所有橋的最小邊權。若不存在橋，輸出-1。

題解：

這個就直接dfs一下，利用時間戳來求橋就行了。但是註意如果邊權為0的時候，答案為1。。因為至少需要一個人去搬炸彈。

代碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 1005;
int n,m,tot;
int head[N];
struct Edge{
    int u,v,next,w;
}e[N*N<<1];
int T;
int dfn[N],low[N],cut[N*N];
void adde(int u,int v,int w){
    e[tot].u=u;e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;
}
void init(){
    T=0;tot=0;
    memset(head,-1,sizeof(head));
    memset(cut,0,sizeof(cut));
    memset(dfn,0,sizeof(dfn));
}
void Tarjan(int u,int pre){
    dfn[u]=low[u]=++T;
    int k = 0;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(v==pre &&!k){
            k=1;
            continue ;
        }
        if(!dfn[v]){
            Tarjan(v,u);
            low[u]=min(low[u],low[v]);
        }else{
            low[u]=min(low[u],dfn[v]);
        }
        if(low[v]>dfn[u]){
            cut[i]=cut[i^1]=1;
        }
    }
}int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n+m<=0) break;
        init();
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            adde(u,v,w);adde(v,u,w);
        }
        int cnt = 0;
        for(int i=1;i<=n;i++){
            if(!dfn[i]){
                Tarjan(i,i);
                cnt++;
            }
        }
        if(cnt>1){
            cout<<0<<endl;
            continue ;
        }
        int ans = 0x3f3f3f3f;
        for(int i=0;i<tot;i+=2){
            if(cut[i])ans=min(ans,e[i].w);
        }
        if(ans==0x3f3f3f3f) puts("-1");
        else cout<<max(ans,1)<<endl;
    }
    return 0;
}

HDU4738:Caocao's Bridges(求橋)