HDU 1102 Constructing Roads （最小生成树）

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19673 Accepted Submission(s): 7510


Problem 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
3
0 990 692
990 0 179
692 179 0
1
1 2


Sample Output

179

题解：建路使各个村庄都连通N个村庄，

0 990 692  ····························1—（0）—1，1——（990）——2，1——(692)——3
990 0 179 ························· 2——（990）——1,  2——（0）————2，2———（179）——3

692 179 0·································3——（692）——1，3——（179）——2，3———（0）———3

1            ·························已经建好的路数
1 2         ···························路（可以将建好的路之间的距离定为0）

采用prime 算法

代码：

///#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;const int maxx=105;const int inf=0x3f3f3f3f;int n,ans;int book[maxx];///记录int e[maxx][maxx];int dis[maxx];///存放最短距离///prime算法核心void prime (){     for(int i=1; i<=n; i++)///看似1-其他点的距离    {        dis[i]=e[1][i];        book[i]=0;    }    book[1]=1;    dis[1]=0;    int u;    for(int i=1; i<=n; i++)    {        int minn=inf;        for(int j=1; j<=n; j++)            if(!book[j]&&dis[j]<minn)            {                minn=dis[j];                u=j;            }        if(minn==inf)            break;        book[u]=1;        ans+=dis[u];        for(int v=1; v<=n; v++)        {            if(!book[v]&&dis[v]>e[u][v])                dis[v]=e[u][v];        }    }}int main(){    while(~scanf("%d",&n))    {        for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            scanf("%d",&e[i][j]);        int q;        scanf("%d",&q);        while(q--)        {            int a,b;            scanf("%d %d",&a,&b);            e[a][b]=e[b][a]=0;///将已经建好的路之间的距离定为0        }        ans=0;        prime();        printf("%d\n",ans);    }    return 0;}