Constructing Roads
Time Limit:2000MSMemory Limit:65536K
Total Submissions:20643Accepted:8697
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
题意：
首先给一个数字N，，代表多少个村庄，然后就是N*N的矩阵，代表两两村庄的距离。再给一个数Q，接下来Q行：a，b表示第a个与第b个村庄已经相连。求全部村庄联通情况下还需要连接的最短路长。
思路：
很明显的kruskal算法的应用。不过让我想郁闷的是因为我从早上写到晚上，到了晚上十点多有点累就不愿看英文题，直接找题解了解题意，结果别人题意解释的好复杂，我看完好像抽人。。。
代码：
//252K63MS#include<iostream>#include<algorithm>#include<cstdio>using namespace std;#define inf 0xffffffftypedef struct node{int l,r;int w;bool operator < (const node &a) const{return w < a.w;}}node;node p[10010];int pre[110];int n,m;int num;//边的个数int Find(int x){return pre[x]==x?pre[x]:Find(pre[x]);}int Union(int a,int b){int f1=Find(a),f2=Find(b);if(f1!=f2){pre[f1]=f2;return 1;}return 0;}void kruskal(){int sum=0;for(int i=0;i<num;i++){if(!Union(p[i].l,p[i].r))continue;sum+=p[i].w;}printf("%d\n",sum);}int main(){scanf("%d",&n);num=0;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){p[num].l=i;p[num].r=j;scanf("%d",&p[num++].w);}}for(int i=0;i<=n;i++)pre[i]=i;scanf("%d",&m);int a,b;int f1,f2;while(m–){scanf("%d%d",&a,&b);Union(a,b);}sort(p,p+num);kruskal();return 0;}
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接受失败，是我们不常听到或看到的一个命题，
