POJ - 2421 Constructing Roads

Constructing Roads

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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

今天kruskal了好几题了，难道要用到kruskal的就没有难题吗？

#include<iostream>#include<algorithm>using namespace std;const int mnx = 105;const int mxe = 104 / 2 * 105;struct edge{int u, v, c;bool operator<(const edge&b) const{return c < b.c;}}e[mxe];int map[mnx][mnx];int n, m;int F[mnx];int find(int x){if (x != F[x]) F[x] = find(F[x]);return F[x];}int kruskal(){sort(e + 1, e + m + 1);for (int i = 1; i <= n; i++) F[i] = i;int ret = 0;for (int i = 1; i <= m; i++){int u = e[i].u, v = e[i].v, c = e[i].c;u = find(u), v = find(v);if (u != v){ret += c;F[u] = v;}}return ret;}int main(){while (scanf("%d", &n) != EOF){for (int i = 1; i < n; i++)F[i] = i;int jishu = 1;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++){cin >> map[i][j];}int q;cin >> q;int a, b;for (int i = 0; i < q; i++){cin >> a >> b;map[a][b] = map[b][a] = 0;     //对于已经存在的边，边权改为0.}for (int i = 1; i <= n;i++)for (int j = i + 1; j <= n; j++){e[jishu].u = i;e[jishu].v = j;e[jishu].c = map[i][j];jishu++;}m = jishu - 1;int ans=kruskal();cout << ans << endl;}}