HDU3448 Bag Problem 01思想+深搜

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3448

题目大意：
Problem Description
0/1 bag problem should sound familiar to everybody. Every earth man knows it well. Here is a mutant: given the capacity of a bag, that is to say, the number of goods the bag can carry (has nothing to do with the volume of the goods), and the weight it can carry. Given the weight of all goods, write a program that can output, under the limit in the above statements, the highest weight.
 
Input
Input will consist of multiple test cases The first line will contain two integers n (n<=40) and m, indicating the number of goods and the weight it can carry. Then follows a number k, indicating the number of goods, k <=40. Then k line follows, indicating the weight of each goods The parameters that haven’t been mentioned specifically fall into the range of 1 to 1000000000.


Output
For each test case, you should output a single number indicating the highest weight that can be put in the bag.


Sample Input
5 100
8
8 64 17 23 91 32 17 12
5 10
3
99 99 99


Sample Output
99
0

#include <iostream>#include <algorithm>using namespace std;int n,m,k;__int64 Max;int a[50];int visited[50];bool cmp(int a, int b) {    return a<b;}void DFS(int s, int N, int M) {    if(s >= k || N > n) {        return;    }    if(Max < M) {        Max = M;    }    for(int i = s + 1; i < k; i++) {        if(visited[i] == 0) {            if(M + a[i] > m) {                continue;            }            visited[i] = 1;            DFS(i, N + 1, M + a[i]);            visited[i] = 0;        }    }}int main() {    while(cin>>n>>m) {        cin>>k;        int i;        for(i = 0; i < k; i++) {            cin>>a[i];            visited[i] = 0;        }        sort(a, a + k, cmp);        __int64 r = 0;        for(i = k - 1; i > k - 1 - n; i--) {            r += a[i];        }        if(r <= m) {            cout<<r<<endl;        } else {            Max = 0;            DFS(-1, 0, 0);            cout<<Max<<endl;        }    }    return 0;}