NYOJ 927 The partial sum problem 【DFS】+【剪枝】

The partial sum problem

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

描写叙述

One day,Tom’s girlfriend give him an array A which contains N integers and asked him:Can you choose some integers from the N integers and the sum of them is equal to K. 

输入

There are multiple test cases.
Each test case contains three lines.The first line is an integer N(1≤N≤20),represents the array contains N integers. The second line contains N integers,the ith integer represents A[i](-10^8≤A[i]≤10^8).The third line contains an integer K(-10^8≤K≤10^8).

输出

If Tom can choose some integers from the array and their them is K,printf ”Of course,I can!”; other printf ”Sorry,I can’t!”.

例子输入

4
1 2 4 7
13
4
1 2 4 7
15

例子输出

Of course,I can!
Sorry,I can't!

这题非常经典，剪枝的时候要细心。

#include <stdio.h>
#include <stdlib.h>
int n, arr[22], sum, vis[22], ok, count;
const char *sam[] = {"Sorry,I can't!
", "Of course,I can!
"};

int cmp(const void *a, const void *b){
	return *(int *)a - *(int *)b;
}

void DFS(int k){
	if(count == sum){
		ok = 1; return;
	}
	
	for(int i = k; i < n; ++i){
		if(i && arr[i] == arr[i-1] && !vis[i-1]) //cut
			continue;
		if(count > sum && arr[i] > 0) return; //cut
		
		count += arr[i]; vis[i] = 1;
		DFS(i + 1);
		if(ok) return;
		count -= arr[i]; vis[i] = 0;
	}
}

int main(){
	while(scanf("%d", &n) == 1){
		for(int i = 0; i < n; ++i){
			scanf("%d", arr + i);
			vis[i] = 0;
		}
		scanf("%d", &sum);
		qsort(arr, n, sizeof(int), cmp);
		count = ok = 0; DFS(0);
		printf(ok ? sam[1] : sam[0]);
	}
	return 0;
}