POJ 2253 Prim算法及测试数据

prim算法求最小生成树中的最大边

测试数据：

2
0 0
3 4

3
17 4
19 4
18 5
8
1 1
4 0
1 2
2 2
3 2
4 2
3 0
5 1
3
9 10
10 10
100 10
6
5 5
100 100
4 4
3 3
2 2
1 1
5
1 2
2 1
3 2
4 1
5 2
3
999 999
1 1
3 3
0

结果

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Scenario #3
Frog Distance = 1.414

Scenario #4
Frog Distance = 1.000

Scenario #5
Frog Distance = 134.350

Scenario #6
Frog Distance = 1.414

Scenario #7
Frog Distance = 1408.557

#include <iostream>
#include <math.h>
#define MAX 201

#define sq(a)  ((a)*(a)) 
float dis[MAX][MAX];
typedef struct node{
    int x,y;
    
}Node;
long int INF =99999999;
bool inTree[MAX];
Node nodes[MAX];
void ptable(int count){
    int i,j;
    for(i=0;i<count;i++){
        for(j=0;j<count;j++)
            printf("(%d,%d)%10f",i,j,dis[i][j]);
        printf("\n");
    }
}
bool allInTree(int count){

    for(int i=0;i<count;i++)
        if(!inTree[i])return false;
    return true;
}
int prim(int v,int n){
    int i,j,ti,tj;
    float mindis,maxdis = -1;
    memset(inTree,false,sizeof(inTree));
    inTree[0] = true;
    while(!allInTree(n)){
        mindis=INF;

        //找到最小权值
        for(i=0;i<n;i++)
        {
            if(inTree[i]){
                for(j=0;j<n;j++)
                    if(i!=j&&!inTree[j]&&dis[i][j]<=mindis){
                        mindis=dis[i][j];
                        ti = i;
                        tj = j;
                    }
            }
        }
        if(mindis>maxdis)maxdis=mindis;
        //找到终点
        if(ti==1||tj==1){
            
            return maxdis;
        }
        //printf("%d %d\nIntree:",ti,tj);
        //inTree[ti]=true;
        inTree[tj]=true;

        //for(int k=0;k<n;k++)
            //if(inTree[k])printf("%d ",k);
            //printf("\n");
        //ptable(n);
    }
    
        
        

    return maxdis;
}
int main(int argc, char* argv[])
{
    //freopen("i://t.txt","r",stdin);
    //freopen("i://out.txt","w",stdout);
    int count,i,j,c = 1;
    while(scanf("%d",&count)&&count){
    
    
    
    for(i=0;i<count;i++)
    {
        scanf("%d%d",&nodes[i].x,&nodes[i].y);
        for(j=0;j<i;j++){
            dis[i][j]=sq(nodes[i].x-nodes[j].x)+sq(nodes[i].y-nodes[j].y);
            dis[j][i]=dis[i][j];
        }
        dis[i][i] = 0;
    }    
    printf("Scenario #%d\n",c++);
    printf("Frog Distance = %.3lf\n\n",sqrt((float)prim(0,count)));
    


    
    }
    return 0;
}