HDU 3371Connect the Cities(最小生成树)

最新推荐文章于 2018-10-08 00:59:02 发布

原创最新推荐文章于 2018-10-08 00:59:02 发布 · 299 阅读

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本文探讨了在2100年海平面上升导致多数城市消失的背景下，如何通过建造最少数量的道路来重新连接剩余的城市，同时确保成本最低。文章提供了一个算法解决方案，并详细解释了其工作原理。

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Connect the Cities
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18982    Accepted Submission(s): 4609Problem Description
In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  
 
Input
The first line contains the number of test cases.Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.To make it easy, the cities are signed from 1 to n.Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
 
Output
For each case, output the least money you need to take, if it’s impossible, just output -1. 
 
Sample Input
1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

Sample Output

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int per[505];
struct money{
	int x,y,z;
}my[25005];
int find(int x){
	int r=x;
    while(per[r]!=r) r=per[r]; 
	int i=x;int j;//路径压缩 
	while(i!=r){
		j=per[i];
		per[i]=r;
		i=j;
	} 
	return r;
}
void join(int r1,int r2){
	int f1=find(r1);
	int f2=find(r2);
	if(f1!=f2) per[f1]=f2;
}
bool cmp(money m1,money m2){
	return m1.z<m2.z;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n,m,k;
		bool flag=true;
		int minmoney=0;
		scanf("%d%d%d",&n,&m,&k); 
		for(int i=0;i<m;i++){
			scanf("%d%d%d",&my[i].x,&my[i].y,&my[i].z);
		}
		for(int i=1;i<=n;i++){
			per[i]=i;
		} 
		for(int i=0;i<k;i++){
			int num;scanf("%d",&num);
		    int first;scanf("%d",&first);//输入第一个 以便于合并 
			for(int j=0;j<num-1;j++){
				int l;scanf("%d",&l);
				join(first,l);//合并 
			}
		}
		sort(my,my+m,cmp);
		int l=0;
		for(int i=1;i<=n;i++){
			if(per[i]==i) l++;//判断还有几个独立的子树 
		}
		for(int i=0;i<m;i++){
			if(find(my[i].x)!=find(my[i].y)){
				join(my[i].x,my[i].y);
				minmoney+=my[i].z;
				l--;
			}
			if(l==1){
				flag=true;break;
			} 
		}
		
		
		if(flag==true)	printf("%d\n",minmoney);
		else printf("-1\n");
	}
	return 0;
}