• 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/chenjianxiang/p/3544452.html
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