POJ - 1664 放苹果(递归)

题意：把M个同样的苹果放在N个同样的盘子里，允许有的盘子空着不放，问共有多少种不同的分法？（用K表示）5，1，1和1，5，1 是同一种分法。

分析：枚举每个盘子可以放的苹果数都是0~M，最后去重就可以了。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int m, n;
vector<int> tmp;
vector<int> a[MAXN];
int cnt;
void dfs(int now, int id, int sum){
    if(sum > m) return;
    if(id == n){
        if(sum == m){
            int t = 0;
            sort(tmp.begin(), tmp.end());
            for(int i = 0; i < cnt; ++i){
                bool ok = false;
                for(int j = 0; j < n; ++j){
                    if(a[i][j] != tmp[j]){
                        ok = true;
                        break;
                    }
                }
                if(ok) ++t;
            }
            if(t == cnt){
                for(int i = 0; i < n; ++i){
                a[cnt].push_back(tmp[i]);
                }
                ++cnt;
            }
        }
        return;
    }
    for(int i = 0; i <= m; ++i){
        tmp.push_back(now + i);
        dfs(now + i, id + 1, sum + now + i);
        vector<int>::iterator it;
        for(it = tmp.begin(); it != tmp.end(); ++it){
            if(*it == now + i){
                tmp.erase(it);
                break;
            }
        }
    }
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        cnt = 0;
        tmp.clear();
        for(int i = 0; i < MAXN; ++i){
            a[i].clear();
        }
        scanf("%d%d", &m, &n);
        dfs(0, 0, 0);
        printf("%d\n", cnt);
    }
    return 0;
}

 

来源：https://www.cnblogs.com/tyty-Somnuspoppy/p/6151551.html