POJ

Constructing Roads

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 24278 Accepted: 10442

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

PKU Monthly,kicc

给出n个点彼此之间的距离以及已经联通的点的标号，求最小生成树的权和。

刚开始看样例菜题意猜出了完全不一样的题意2333333

在最小生成树的模板下提前对联通的点并查集就好了。。感觉可以写一写克鲁斯卡尔的教学博客了？

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<set>#include<queue>#include<map>#include<stack>#define LL long long#define max3(x, y, z) max((x), max((y), (z)))#define min3(x, y, z) min((x), min((y), (z)))#define pb push_back#define mp make_pairusing namespace std;const int N = 100010;struct Edge{    int u;    int v;    int w;}edges[N];bool cmp(Edge a, Edge b){    return a.w < b.w;}int n, cnt;int ne[N];int ans;void ini(){    for(int i = 1; i <= n; i ++){        ne[i] = i;    }}int fi(int x){    int t = x;    while(t != ne[t]){        t = ne[t];    }    while(x != t){        int q = ne[x];        ne[x] = t;        x = q;    }    return t;}void join(int x, int y){    ne[x] = y;}LL Kru(){    int num = 0;    LL ans = 0;    sort(edges, edges + cnt, cmp);    for(int i = 0; i < cnt; i ++){        int a = fi(edges[i].u);        int b = fi(edges[i].v);        if(a != b){            join(a, b);            ans += edges[i].w;            num ++;        }        if(num == cnt - 1) break;    }    return ans;}int main(){    int m;    while(scanf("%d", &n) == 1){        cnt = 0;        for(int i = 1; i <= n; i ++){            for(int j = 1; j <= n; j ++){                int w;                scanf("%d", &w);                if(i != j){                    edges[cnt].u = i;                    edges[cnt].v = j;                    edges[cnt].w = w;                    cnt ++;                }            }        }        ini();        scanf("%d", &m);        for(int i = 0; i < m; i ++){            int u, v;            scanf("%d%d", &u, &v);            int a = fi(u);            int b = fi(v);            if(a != b){                join(a, b);            }        }        printf("%lld\n", Kru());    }    return 0;}