Hdu1102 constructing roads [Minimum Spanning Tree prim]

Constructing roads

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 13756 accepted submission (s): 5223

Problem descriptionthere are n versions, which are numbered from 1 to n, And You shoshould build some roads such that every two versions ages 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 versions ages and your job is the build some roads such that all the versions are connect and the length of all the roads built is minimum.
 

 

Inputthe first line is an integer N (3 <= n <= 100), which is the number of ages. 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 shocould 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.
 

 

Outputyou shoshould output a line contains an integer, which is the length of all the roads to be built such that all the versions are connected, and this value is minimum.
 

 

Sample Input

30 990 692990 0 179692 179 011 2

 

 

Sample output

179

The point of this question is that the question is actually multi-group data...

 

#include <stdio.h>#include <string.h>#define maxn 102int map[maxn][maxn];bool vis[maxn];void Prim(int n){int i, j, len = 0, count = 0, tmp, u;vis[1] = 1;while(count < n - 1){for(i = 1, tmp = -1; i <= n; ++i){if(!vis[i]) continue;for(j = 1; j <= n; ++j)if(!vis[j] && (map[i][j] < tmp || tmp == -1)){tmp = map[i][j]; u = j;}}if(tmp != -1){++count;vis[u] = 1;len += tmp;}}printf("%d\n", len);}int main(){//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int n, q, a, b, i, j;while(scanf("%d", &n) != EOF){memset(vis, 0, sizeof(vis));for(i = 1; i <= n; ++i)for(j = 1; j <= n; ++j)scanf("%d", &map[i][j]);scanf("%d", &q);while(q--){scanf("%d%d", &a, &b);map[a][b] = map[b][a] = 0;}Prim(n);}return 0;}