hdu 1518 Square (dfs搜索可参考poj1011)
2014-08-02 16:12
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Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8589 Accepted Submission(s): 2784
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
#include"stdio.h" #include"string.h" #include"queue" #include"vector" #include"algorithm" using namespace std; #define N 25 #define max(a,b) (a>b?a:b) int a ,n,t,mark ,sum; bool cmp(int a,int b) { return a>b; } int dfs(int index,int len,int rest) { int i; if(rest==t) return 1; for(i=index;i<n;i++) { if(!mark[i]&&a[i]<=len) { mark[i]=1; if(len==a[i]) { if(dfs(0,t,rest-a[i])) return 1; } else if(dfs(i+1,len-a[i],rest-a[i])) return 1; mark[i]=0; //若当前值不能满足条件,进行回溯 if(rest==sum) //若满足此条件,则该条边不能加入任一条边 return 0; if(len==t) //同上 return 0; while(a[i]==a[i+1]) //这条边不行,则和它同长的边也不行 i++; } } return 0; } int main() { int T,i; scanf("%d",&T); while(T--) { scanf("%d",&n); sum=0; for(i=0;i<n;i++) { scanf("%d",&a[i]); sum+=a[i]; } sort(a,a+n,cmp); if(sum%4) printf("no\n"); else { memset(mark,0,sizeof(mark)); if(dfs(0,t=sum/4,sum)) printf("yes\n"); else printf("no\n"); } } return 0; }
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