TOJ 4394 Rebuild Road

描述

    Once,in a kingdom,there are N cities.M roads have been buit such that from one city you can reach any other cities.Between any two cities there is at most one road.But after a series war,D road are destoryed.The king wants to repair the road system.There are two important cities A and B,The king wants to make  the two cities connected as soon as possible.Now it is your job to repair some roads such that A and B are connected and the length of all roads repaired is minimun.

输入

Input may contain seveal test data sets.
For each data set, the first line contain an integer N(2<N<100),indicating the number of cities.The cities are numbered from 1 To N.then the second line contains an integer M ( N-1<M<N*(N-1)/2),indicating the number of roads.Next come M lines,each contains three integers I,J,K,which means that there is a road between I and J,whose length is K.
 Then next line contains an integer D(1<=D<=M),indicating the number of roads which are destoryed.The following D lines each contain two integer I,J,which means that the road which directly connected city I and J has been destoryed.
 The last line contains two integer A,B,indicating the two important cities.

Input is ended by N = 0,which should not be processed.

输出

For each test case,just output one line with the total length of the roads needed to repair such that A and B are connected.If we could not raech B from A,output"-1".

样例输入

3
2
1 2 1
2 3 2
1
1 2
1 3
0

样例输出

1

每个城市之间都有一些路连接，某些城市之间的有遭到了破坏，现在要求我们求修路的最短长度。

如果修完所有的路也无法保证通行则输出-1。

逆向思维，让城市之间已经存在的路的长度为0，需要修补的初始化要修补的路的长度。这样就变成了简单的单源最短路径问题了。

#include <stdio.h>
#include <string.h>
#define MAXN 110
#define inf 0x3f3f3f3f

int N;
int m1[MAXN][MAXN];
int m2[MAXN][MAXN];
int dist[MAXN];
int visited[MAXN];

void dijkstra(int begin){
	int i,j,k;
	k=begin;
	for(j=1; j<=N; j++){		
		visited[j]=0;	
		if(j!=k){			
			dist[j]=m2[k][j];			
		}	
	}
	visited[k]=1;	
	for(i=1;i<N-1;i++){
		int min=inf;
		k=0;	
		for(j=1; j<=N; j++){			
			if(!visited[j]&&dist[j]<min){
				min=dist[j];
				k=j;
			}		
		}
		if(k==0)break;
		visited[k]=1;		
		for(j=1; j<=N; j++){		
			if(!visited[j]&& dist[k]+m2[k][j]<dist[j]){			
				dist[j]=dist[k]+m2[k][j];		
			}
		}		
	}	
}

int main(){	
	int M,D,a,b,c;
	int B,E;
	while( scanf("%d",&N)!=EOF && N){			
		for( int i=1 ;i<=N; i++ ){		
			for( int j=1; j<=N; j++ ){
				if(i==j)m2[i][j]=0;			
				else m2[i][j]=inf;		
			}		
		}
		scanf("%d",&M);	
		for( int i=0; i<M; i++ ){			
			scanf("%d%d%d",&a,&b,&c);
			m1[a][b]=c;
			m1[b][a]=c;
			m2[a][b]=0;
			m2[b][a]=0;
		}		
		scanf("%d",&D);
		for( int i=0; i<D; i++ ){			
			scanf("%d%d",&a,&b);
			m2[a][b]=m1[a][b];
			m2[b][a]=m1[b][a];			
		}
		scanf("%d%d",&B,&E);		
		dijkstra(B);		
		if(dist[E]==inf){
			printf("-1
");
		}else{
			printf("%d
",dist[E]);
		}
	}	
	return 0;
}