Constructing RoadsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15230Accepted Submission(s): 5826
Problem Description
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
Sample Input
30 990 692990 0 179692 179 011 2
Sample Output
179
Source
kicc
只要把已经建好的路的权值设置为0就可以了。
水题，没什么好说的
代码：
#include <stdio.h>#include <string.h>#define INF 100000000#define MAX 110int graph[MAX][MAX] ;int prim(int n){int lowCost[MAX],closest[MAX];bool visited[MAX];memset(visited,false,sizeof(visited)) ;for(int i = 1 ; i <= n ; ++i){lowCost[i] = graph[1][i] ;closest[i] = 1 ;}visited[1] = true ;int sum = 0 ;for(int i = 0 ; i < n-1 ; ++i){int min = INF , index = -1 ;for(int j = 1 ; j <= n ; ++j){if(!visited[j] && min>lowCost[j]){index = j ;min = lowCost[j] ;}}if(index == -1){break ;}sum += lowCost[index] ;visited[index] = true ;for(int j = 1 ; j <= n ; ++j){if(!visited[j] && lowCost[j]>graph[index][j]){lowCost[j] = graph[index][j] ;closest[j] = index ;}}}return sum ;}int main(){int n ;while(~scanf("%d",&n)){for(int i = 1 ; i <= n ; ++i){for(int j = 1 ; j <= n ; ++j){int w ;scanf("%d",&w) ;graph[i][j] = w ;}}int q ;scanf("%d",&q);for(int i = 0 ; i < q ; ++i){int x , y ;scanf("%d%d",&x,&y);graph[x][y] = graph[y][x] = 0 ;}int sum = prim(n) ;printf("%d\n",sum) ;}return 0 ;}与君共勉
，从起点，到尽头，也许快乐，或有时孤独，
