HDU - 4738 Caocao's Bridges(桥)

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题目大意：有N座小岛，M座桥，每座桥都有相应的士兵保守。现在你有一单位的炸药可以炸掉一座桥。要炸掉一座桥的话，你派的士兵数量至少要等于该桥的士兵数量
你的任务是使这些岛屿连通。问需要破坏哪一座桥，且派的士兵最小

解题思路：很简单吧，水题，找最小权值的“桥“就可以了
但是坑点多啊
如果不连通的话，就不用派士兵了
如果桥的最小权值是0的话，那么就是1了，因为至少得派一个人

#include <cstdio>#include <cstring>#include <vector>#include <stack>using namespace std;#define max(a,b)((a)>(b)?(a):(b))#define min(a,b)((a)<(b)?(a):(b))const int MAXNODE = 1005;const int MAXEDGE = 2000010;const int INF = 0x3f3f3f3f;struct Edge{    int v, id, next, cost;    Edge() {}    Edge(int v, int id, int next, int cost): v(v), id(id), next(next), cost(cost){}}E[MAXEDGE * 2];int head[MAXNODE], pre[MAXNODE], lowlink[MAXNODE];int n, m, tot, dfs_clock; bool flag;int ans, cnt;void AddEdge(int u, int v, int cost, int id) {    E[tot] = Edge(v, id, head[u], cost);    head[u] = tot++;    u = u ^ v; v = u ^ v; u = u ^ v;    E[tot] = Edge(v, id, head[u], cost);    head[u] = tot++;}void init() {    memset(head, -1, sizeof(head));    tot = 0;    int u, v, c;    for (int i = 1; i <= m; i++) {        scanf("%d%d%d", &u, &v, &c);        AddEdge(u, v, c, i);    }}void dfs(int u, int fa) {    pre[u] = lowlink[u] = ++dfs_clock;    cnt++;    for (int i = head[u]; ~i; i = E[i].next) {        int v = E[i].v;        if (E[i].id == fa) continue;        if (!pre[v]) {            dfs(v, E[i].id);            lowlink[u] = min(lowlink[u], lowlink[v]);//更新            if (lowlink[v] > pre[u]) {                 flag = true;                ans = min(ans, E[i].cost);            }        }        else lowlink[u] = min(lowlink[u], pre[v]);    }}void find_bcc() {    memset(pre, 0, sizeof(pre));    dfs_clock = 0;    dfs(1, -1);}void solve() {    flag = false; ans = INF;    cnt = 0;    find_bcc();    if (cnt != n) {        printf("0\n");         return ;    }    if (!flag)        printf("-1\n");    else        printf("%d\n", ans == 0 ? ans + 1 : ans);}int main() {    while (scanf("%d%d", &n, &m) != EOF && n + m) {        init();        solve();    }    return 0;}

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