poj 2362 Square

Square

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 15977 Accepted: 5505

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

Waterloo local 2002.09.21

限制条件还是比较多的，首先和能够被4整出得到side，然后最长的棍子不可以超过sum的1/4 。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>using namespace std;int n,len[30],side;bool vis[30],flag;inline bool cmp(int x,int y){    return x>y;}void read(){    memset(vis,0,sizeof(vis));    scanf("%d",&n);    for(int i=1;i<=n;i++)    scanf("%d",&len[i]);}void dfs(int num,int length,int cs){    if(cs==3) //形成三边的时候就形成square了    {        flag=1;        return ;    }    if(length==side) dfs(1,0,cs+1);    if(flag)return ;    for(int i=num;i<=n;i++)    {        if(!vis[i]&&length+len[i]<=side)        {            vis[i]=true;            dfs(i+1,len[i]+length,cs);//尝试一次一次的加边使之达到side            if(flag) return ;            vis[i]=false;        }    }}void solve(){    flag=0;    dfs(1,0,0);    if(!flag)    {        printf("no\n");        return;    }    printf("yes\n");}int main(){    int T;    scanf("%d",&T);    while(T--)    {        read();        int sum=0,maxx=0;        for(int i=1;i<=n;i++)        {            sum+=len[i];            maxx=max(maxx,len[i]);        }        if(sum%4!=0||4*maxx>sum)        printf("no\n");        else        {            sort(len+1,len+n+1,cmp);            side=sum/4;            solve();        }    }    return 0;}

看了下，sort函数里面的形式，还有要定义cmp。。。第一次用这个的说