HDU1518 Square (DFS)
2017-02-11 23:03
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Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14057 Accepted Submission(s): 4439
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
University of Waterloo Local Contest 2002.09.21
思路:一道基础的DFS题目,注意要剪枝,在进行搜索时,如果某一条边还没有搜索完毕,则下次搜索应该从数组a的下一个元素开始,因为在这之前的所有元素都已经搜索过了,再搜索就重复了。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int a[25],vis[25],flag;
void dfs(int k,int cnt, int cur,int aver, int m){
if(cnt == 3){
flag = 1;
}
if(flag)
return;
for(int i = k; i < m; i ++){
if(!vis[i]){
if(cur + a[i] < aver){
vis[i] = 1;
dfs(i + 1,cnt,cur+a[i],aver,m);
vis[i] = 0;
}
else if(cur + a[i] == aver){
vis[i] = 1;
dfs(0,cnt+1,0,aver,m);
vis[i] = 0;
}
}
}
}
int main(){
int n,m;
scanf("%d",&n);
while(n --){
scanf("%d",&m);
int sum = 0;
for(int i = 0; i < m; i ++){
scanf("%d",&a[i]);
sum += a[i];
}
if(sum % 4){
printf("no\n");
continue;
}
flag = 0;
for(int i = 0; i < m; i ++){
if(a[i] > sum / 4){
flag = 1;
break;
}
}
if(flag){
printf("no\n");
continue;
}
flag = 0;
memset(vis,0,sizeof(vis));
dfs(0,0,0,sum/4,m);
if(flag)
printf("yes\n");
else
printf("no\n");
}
return 0;
}
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