HDU4738 Caocao's Bridges

无向图有重边的tarjan算法 
接收数据是把每条边的权值保存起来 
有重边的tarjan算法要判断走重边时的low值 
然后找出最小权值的桥 
另外三个注意点： 
1、原本不是一个强联通图的话，派的人是0 
2、如果不存在桥的话，输出-1 
3、如果桥的最小权值是0，则至少派一个人去炸桥

#include<stdio.h>
#include<string.h>
#include<algorithm>

using namespace std;

const int maxn = 1005;
const int maxm = 1005*1005;
int n, m;
int top, tol, cnt;

struct Node {
    int v;
    int next;
    int w;
};
Node node[maxm * 2];
int dfn[maxn];
int low[maxn];
int head[maxn];

void init() {
    tol = top = cnt = 0;
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(node, 0, sizeof(node));
    memset(head, -1, sizeof(head));
}

void addnode(int u, int v, int w) {
    node[tol].v = v;
    node[tol].w = w;
    node[tol].next = head[u];
    head[u] = tol++;
}

void dfs(int u, int fa) {
    dfn[u] = low[u] = ++cnt;
    int cnt = 0;
    for(int i=head[u]; i!=-1; i=node[i].next) {
        int v = node[i].v;
        if(!dfn[v]) {
            dfs(v, u);
            low[u] = min(low[u], low[v]);
        } else if(v == fa) {
            if(cnt) 
                low[u] = min(low[u], dfn[v]);
            cnt++;
        } else
            low[u] = min(low[u], dfn[v]);
    }
}

int tarjan() {
    int time = 0;
    for(int i=1; i<=n; i++) {
        if(!dfn[i]) {
            dfs(i, i);
            time++;
        }
    }
    return time;
}

int solve() {
    int ans = 0x3f3f3f3f;
    for(int u=1; u<=n; u++) {
        for(int i=head[u]; i!=-1; i=node[i].next) {
            int v = node[i].v;
            if(dfn[u] < low[v] && u != v) {
                ans = min(ans, node[i].w);
            }
        }
    }
    return ans;
}

int main() {
    while(scanf("%d%d",&n, &m), n||m) {
        init();
        for(int i=1; i<=m; i++) {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            addnode(u, v, w);
            addnode(v, u, w);
        }
        int time = tarjan();
        if(time > 1)
            printf("0\n");
        else {
            int ans = solve();
            if(ans == 0x3f3f3f3f)
                printf("-1\n");
            else if(ans == 0)
                printf("1\n");
            else
                printf("%d\n", ans);    
        }
    }
    return 0;
}

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