TOJ 2523 ：最大生成树

2523: Heavy Transportation 分享至QQ空间
时间限制(普通/Java):3000MS/30000MS 内存限制:65536KByte
总提交: 163 测试通过:48
描述

Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.

Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo’s place) to crossing n (the customer’s place). You may assume that there is at least one path. All streets can be travelled in both directions.

输入

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

输出

The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

样例输入

1
3 3
1 2 3
1 3 4
2 3 5

样例输出

Scenario #1:
4
题目来源
题解：题意是每条路都有载重的限制，求一条从1出发到N最大的最小载重的路径。这题可以用最大生成树用着最短路做，最短路再补~
最大生成树：
要注意退出的条件不是图已经全部联通，而是1点和N点联通。

最大生成树

#include
using namespace std;
const int maxn = 1e5 + 5,maxN = 1e3 + 5;
struct edge{ //结构体存边
	int x,y,v;
}e[maxn];
int pre[maxN];
bool cmp(edge a,edge b){  //大到小排序
	return a.v>b.v;
}
int getfather(int x){ //并查集模板找父亲
	return x==pre[x]?x:pre[x]=getfather(pre[x]);
}
int main(){
	int t,n,m,Case=0;
	ios::sync_with_stdio(false); 
	cin >> t;
	while(t--){
		cin >> n >> m;
		int cnt=0;
		for(int i=1;i<=n;i++) pre[i]=i;
		for(int i=0;i<m;i++) cin >> e[i].x >> e[i].y >> e[i].v;
		sort(e,e+m,cmp);
		for(int i=0;i<m;i++){
			int fx=getfather(e[i].x);
			int fy=getfather(e[i].y);
			if(fx!=fy){  //判断这两个点是否联通，不联通增加新边
                pre[fx]=fy;
                ++cnt;
			}
			int f0=getfather(1),fn=getfather(n);
			if(cnt==n-1||f0==fn){  //判断1点和N点是否联通
                printf("Scenario #%d:\n%d\n\n",++Case,e[i].v);
                break;
			}
		}
	}
}