H - Constructing Roads（解题报告）

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

这道题是一个最小生成树的问题，有多种解法，我在这里就用prime解法来解决。这里先定义一个二维数组存储各目的地之间的距离，然后在处理其中小的部分提取即可。

下面是代码：

#include <stdio.h>#include <string.h>#include <map>using namespace std;#define MAX 0x3f3f3f3fint main(){map<int,int>letre[1005];int N,Q,i,j,con1,con2,vel[105],cnt,min,res,judge,dis[105];memset(vel,0,sizeof(vel));while(scanf("%d",&N)!=EOF){for(i=0;i<N;i++){for(j=0;j<N;j++){scanf("%d",&letre[i][j]);//初始化，让输入每个目的地对应的距离}}scanf("%d",&Q);while(Q--){scanf("%d%d",&con1,&con2);letre[con1-1][con2-1] = 0;letre[con2-1][con1-1] = 0; //已经联通的地方就让其距离为0}for(i=0;i<N;i++){dis[i] = letre[0][i];}vel[0] = 1;res = 0;for(i=1;i<N;i++){min = MAX;for(j=0;j<N;j++){if(!vel[j]&&min>dis[j]){min = dis[j];judge = j;}} if(min==MAX){break;}res += min;vel[judge] = 1;for(j=0;j<N;j++){if(!vel[j]&&dis[j]>letre[judge][j]){dis[j] = letre[judge][j];//这里的目的是让每一个已经联通的点之间最短距离存入判断的数组中，以便下一次判断}}}printf("%d\n",res);}return 0;}
做完才发现，用map其实就相当于用了二维数组。