poj 2362 Square

来源：互联网发布：java简单请假系统编辑：程序博客网时间：2024/06/14 21:47

题目来源：http://poj.org/problem?id=2362

跟poj1011, nyoj 293大体思路一样！

但是这个题要有自身的剪枝！：http://blog.csdn.net/hearthougan/article/details/22986249

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 25;int Len_Of_Sticks[MAXN];bool visit[MAXN];int num;bool DFS(int CurLen, int kcount, int pos, int EdgeLen){    if(kcount == 3)//有三条边满足，没那么就可以组成正方形，新加判别条件        return true;    for(int i = pos; i < num; ++i)    {        if(!visit[i])        {            if(Len_Of_Sticks[i] + CurLen == EdgeLen)            {                visit[i] = true;                if(DFS(0, kcount+1, 0, EdgeLen))                    return true;                visit[i] = false;                return false;            }            else if(Len_Of_Sticks[i] + CurLen < EdgeLen)            {                visit[i] = true;                if(DFS(Len_Of_Sticks[i] + CurLen, kcount, i+1, EdgeLen))                    return true;                visit[i] = false;                bool flag = (CurLen == 0 ? true : false);                if(flag)                    return false;                while(Len_Of_Sticks[i] == Len_Of_Sticks[i+1])                    ++i;            }        }    }    return false;}int cmp(int a, int b){    return a > b;}int main(){    int T, sum, Edge;    int i;    scanf("%d", &T);    while(T--)    {        scanf("%d", &num);        sum = 0;        for(i = 0; i < num; ++i)        {            scanf("%d", &Len_Of_Sticks[i]);            sum += Len_Of_Sticks[i];        }        if(num < 4 || sum % 4 != 0)//边数小于4 || 长度之和不能被4整除自然不可以        {            printf("no\n");            continue ;        }        memset(visit, false, sizeof(visit));        sort(Len_Of_Sticks, Len_Of_Sticks+num, cmp);        Edge = sum/4;        if(Edge < Len_Of_Sticks[0])//最长的小棒不能大于边！！！        {            printf("no\n");            continue ;        }        if(DFS(0, 0, 0, Edge))            printf("yes\n");        else            printf("no\n");    }    return 0;}

0 0