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

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

 

Sample Output

yesnoyes

 

Source

University of Waterloo Local Contest 2002.09.21

 

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#define maxn 22int 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;}