UVA——1587

问题描述：

解决方案：

#include <iostream>
#include <cstdio>
bool isSame(int edge[6][2])
{
    int a,b;
    int same = 0;
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++){
            if(edge[0][i] == edge[1][j]){
                same = edge[0][i];
                break;
            }
        }
    if(same == 0)
        return false;
    a = edge[0][0];
    if(a == same)
        a = edge[0][1];
    b = edge[1][0];
    if(b == same)
        b = edge[1][1];
    if((a == edge[2][0] && b == edge[2][1]) || (a == edge[2][1] && b == edge[2][0]))
        return true;
    return false;
}
int main()
{
    int edge[6][2];
    int a,b;
    while(scanf("%d",&a) !=EOF){
        scanf("%d",&b);
        if(a > b){
            edge[0][0] = b;
            edge[0][1] = a;
        }else{
            edge[0][0] = a;
            edge[0][1] = b;
        }

        int index = 1;
        int count[6] = {0};
        for(int i = 1;i < 6;i++)
        {
            scanf("%d %d",&a,&b);
            if(a > b){
                int c = a;
                a = b;
                b = c;
            }
            int j;
            for(j = 0; j < index; j++)
                if(edge[j][0] == a && edge[j][1] == b)
                    break;
            if(j >= index)
            {
                edge[index][0] = a;
                edge[index][1] = b;
                index++;
            }else{
                count[j]++;
            }
        }
        bool flag = false;
        if(count[0] == 5 && edge[0][0] == edge[0][1]){
            flag = true;
        }
        for(int i = 0; i < 2; i++)
            if(count[i] == 3 && count[1 - i] == 1 && edge[1 - i][0] == edge[1 - i][1] && (edge[1 - i][0] == edge[i][0] || edge[1 - i][0] == edge[i][1]))
                flag = true;
        if((count[0] == 1 && count[1] == 1 && count[2] == 1 && index == 3 && isSame(edge))|| flag)
            printf("POSSIBLE\n");
        else
            printf("IMPOSSIBLE\n");
    }
    return 0;
}
//核心思路：暴力求解