HDU 4738 Caocao’s Bridges（桥、任何位运算一定都要加括号、因为有重边所以用前向星）
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 )
The input ends with N = 0 and M = 0.
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.
题意：曹操有一个无向图，图中有一些边，周瑜现在要派一些人去炸一条边，使曹操的无向图分为不同部分，曹操的每条边上都有守卫，周瑜派的人不能少于边上守卫数，问最少要派多少人。
- 注意重边处理， 要用链式前向星。
- 守卫为0时，也需要派一个人；
- 如果原本图不连通（tarjan多次），就不需要派人，答案就是0。
- 答案直接输出所有桥中的最小边权。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N = 500007, M = 5000007, INF = 0x3f3f3f3f;
int n, m;
int bridge[N];
int dfn[N], low[N], num;
int head[N], ver[M], nex[M], edge[M], tot;
int ans;
int dcc_cnt;
int min_bridge;
void init()
{
num = tot = dcc_cnt = 0;
memset(head, -1, sizeof head);
memset(dfn, 0, sizeof dfn);
memset(low, 0, sizeof low);
min_bridge = INF;
}
void add(int x, int y, int z)
{
ver[tot] = y;
edge[tot] = z;
nex[tot] = head[x];
head[x] = tot ++ ;
}
void tarjan(int x, int in_edge)
{
dfn[x] = low[x] = ++ num;
for(int i = head[x] ;~i; i = nex[i]){
int y = ver[i], z = edge[i];
if(!dfn[y]){
tarjan(y, i);
low[x] = min(low[x], low[y]);
if(dfn[x] < low[y]){
bridge[i] = bridge[i ^ 1] = true;
min_bridge = min(min_bridge, z);
}
}
else if(i != (in_edge ^ 1))
low[x] = min(low[x], dfn[y]);
}
}
int main()
{
while(~scanf("%d%d", &n, &m) && n ){
init();
for(int i = 1; i <= m; ++ i){
int x, y ,z;
scanf("%d%d%d", &x, &y, &z);
add(x, y, z), add(y, x, z);
}
dcc_cnt = 0;
for(int i = 1; i <= n; ++ i)
if(!dfn[i])
tarjan(i, 0), dcc_cnt ++ ;
if(dcc_cnt > 1){
//tarjan多次说明不连通，直接输出0
printf("0\n");
continue;
}
if(min_bridge == INF)min_bridge = -1;
else if(min_bridge == 0)min_bridge = 1;
printf("%d\n", min_bridge);
}
return 0;
}