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;
}