Hdu2660 Accepted Necklace【简单dfs】

http://acm.hdu.edu.cn/showproblem.php?pid=2660

题目大意：n个珠子里面找k个珠子，使得价值最大但重量不超过某一限定值。

直接对要不要第i颗珠子进行dfs，搜索过程中注意剪枝。

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
using namespace std;
template <class T> void checkmin(T &t,T x) {if(x < t) t = x;}
template <class T> void checkmax(T &t,T x) {if(x > t) t = x;}
template <class T> void _checkmin(T &t,T x) {if(t==-1) t = x; if(x < t) t = x;}
template <class T> void _checkmax(T &t,T x) {if(t==-1) t = x; if(x > t) t = x;}
typedef pair <int,int> PII;
typedef pair <double,double> PDD;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end ; it ++)
const int N = 22;
int a[N] , b[N];
int n , K , ans;
int accept;
void dfs(int id,int vn,int wn,int k) {
    if(id > n || wn > accept) return;
    if(k+n-id+1 < K) return;
    if(id == n + 1) {
        if(k == K) checkmax(ans , vn);
        return;
    }
    if(k == K) checkmax(ans , vn);
    dfs(id+1,vn+a[id],wn+b[id],k+1);
    dfs(id+1,vn,wn,k);
}
int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&K);
        for(int i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]);
        scanf("%d",&accept);
        ans = -1;
        dfs(0,0,0,0);
        printf("%d\n" , ans);
    }
    return 0;
}