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## [$HDU4738$ $Caocao$'$s$ $Bridges$](https://blog.csdn.net/u012587561/article/details/48959779)
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### 一、题目描述
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曹操在长江上建立了一些点，点之间有一些边连着。如果这些点构成的无向图变成了连通图，那么曹操就无敌了。刘备为了防止曹操变得无敌，就打算去摧毁连接曹操的点的桥。但是诸葛亮把所有炸弹都带走了，只留下一枚给刘备。所以刘备只能炸一条桥。
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题目给出$n，m$。表示有$n$个点，$m$条桥。
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接下来的$m$行每行给出$a，b，c$，表示$a$点和$b$点之间有一条桥，而且曹操派了$c$个人去守卫这条桥。
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现在问刘备最少派多少人去炸桥。
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如果无法使曹操的点成为多个连通图，则输出$-1$.
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**测试输入**
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```cpp {.line-numbers}
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3 3
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1 2 7
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2 3 4
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3 1 4
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3 2
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1 2 7
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2 3 4
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0 0
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```
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**测试输出**
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```cpp {.line-numbers}
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-1
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4
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```
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### 二、解题思路
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就是用$tarjan$算法算出桥，比较哪一个的值最小。
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#### $Tips$
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- ①. 有重边，注意重边处理，要用链式前向星。我们一直在用链式前向星，所以这个不是问题，反倒是使用邻接表的同学们需要烦恼了~
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- ②. 如果无向图图本身已经有两个连通图了，就无需派人去炸桥，这时候输出$0$。
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- ③. 如果求出来的最小权值桥的守卫人数为$0$时，也需要派出一个人去炸桥。
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### 三、实现代码
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 500010, M = N << 1;
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const int INF = 0x3f3f3f3f;
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int n, m;
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int res;
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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// Tarjan求割边
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int dfn[N], low[N], ts;
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void tarjan(int u, int fa) {
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dfn[u] = low[u] = ++ts;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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if (!dfn[v]) {
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tarjan(v, u);
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low[u] = min(low[u], low[v]);
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if (low[v] > dfn[u]) res = min(res, w[i]);
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} else
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low[u] = min(low[u], dfn[v]);
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}
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("HDU4738.in", "r", stdin);
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#endif
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while (~scanf("%d%d", &n, &m) && n) {
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ts = idx = 0;
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memset(h, -1, sizeof h);
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memset(dfn, 0, sizeof dfn);
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memset(low, 0, sizeof low);
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res = INF;
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while (m--) {
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int a, b, c;
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scanf("%d%d%d", &a, &b, &c);
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add(a, b, c), add(b, a, c);
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}
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int dcc_cnt = 0;
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i, -1), dcc_cnt++;
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if (dcc_cnt > 1) { // tarjan多次说明不连通，直接输出0
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printf("0\n");
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continue;
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}
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if (res == INF)
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res = -1;
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else if (res == 0)
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res = 1;
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printf("%d\n", res);
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}
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return 0;
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}
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```
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