HDU1518 Square 【剪枝】

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8900    Accepted Submission(s): 2893

Problem Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

 

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

 

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

 

Sample Input

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

 

Sample Output

yes
no
yes

 

Source

field=problem&key=University+of+Waterloo+Local+Contest+2002.09.21&source=1&searchmode=source">University of Waterloo Local Contest 2002.09.21

 

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>

#define maxn 22

int L[maxn], n, tar, times;
bool vis[maxn], ok;

bool DFS(int k, int leftLen) {
    if(!leftLen) {
        if(++times == 4) return true;
        for(int i = 1; i < n; ++i) {
            if(!vis[i]) {
                vis[i] = 1;
                if(DFS(i + 1, tar - L[i]))
                    return true;
                else {
                    --times;
                    vis[i] = 0;
                    return false;
                }
            }
        }
    }

    int i;
    for(i = k; i < n; ++i) {
        if(!vis[i] && L[i] <= leftLen) {
            vis[i] = 1;
            if(L[i-1] == L[i] && !vis[i-1]) {
                vis[i] = 0;
                continue;
            }
            if(DFS(i+1, leftLen - L[i]))
                return true;
            vis[i] = 0;
        }
    }

    return false;
}

int main() {
    int t, i;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        tar = 0;
        for(i = 0; i < n; ++i) {
            scanf("%d", &L[i]);
            vis[i] = 0; tar += L[i];
        }

        if(tar % 4) {
            printf("no\n");
            continue;
        }
        tar /= 4;

        std::sort(L, L + n, std::greater<int>());
        if(L[0] > tar) {
            printf("no\n");
            continue;
        }

        times = 0; vis[0] = 1;
        DFS(1, tar - L[0]);

        printf(times == 4 ? "yes\n" : "no\n");
    }
    return 0;
}