LA4794 Sharing Chocolate

Sharing Chocolate

 给你x*y巧克力，问你能不能切成n块给定面积的巧克力，每次只能沿一条方格线所在直线切且不能同时切割多块巧克力

dp[r][c][S]表示r*c切S集合面积是否可行

不难发现r*c=S集合面积的和是可行的必要条件，因此c可以通过S集合面积的和推出，状态简化为dp[r][S]

枚举S子集S0，把r*c分成r0*c或r*c0分别切S0和S-S0，转移即可

RE一直再想爆栈的事情

结果是内存报了

浪费了一个多小时

mdzz

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;

int n,x,y,a[110],dp[110][1 << 15],sum[1 << 15],ma,ans,t,b[110][1 << 16],num[1 << 15];

int tiaoshi;

int dfs(int r, int S)
{
    if(b[r][S]) return dp[r][S];
    b[r][S] = 1;
    if(num[S] == 1) return dp[r][S] = 1;
    int c = sum[S] / r;
    for(register int SS = (S - 1) & S;SS;SS = (SS - 1) & S)
    {
        if(sum[SS] % r == 0 && ((sum[SS] < r) || (sum[S - SS] < r))) 
            ++ tiaoshi;
        if(sum[SS] % r == 0 && dfs(min(r, sum[SS]/r), SS) && dfs(min(r, (sum[S - SS])/r), S - SS)) 
            return dp[r][S] = 1;    
        if(sum[S - SS]/c == 0)
            ++ tiaoshi;
        if(sum[SS] % c == 0 && ((sum[SS] < c) || (sum[S - SS] < c))) 
            ++ tiaoshi;
        if(sum[SS] % c == 0 && dfs(min(c, sum[SS]/c), SS) && dfs(min(c, (sum[S - SS])/c), S - SS)) 
            return dp[r][S] = 1;
    }
    return dp[r][S] = 0;
}

int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        ++ t;
        read(x), read(y);
        memset(b, 0, sizeof(b));
        memset(sum, 0, sizeof(sum));
        memset(num, 0, sizeof(num));
        for(register int i = 1;i <= n;++ i) read(a[i]);
        ma = 1 << n;
        for(register int S = 0;S < ma;++ S)
            for(register int i = 1;i <= n;++ i)
            {
                if(S == 260) 
                    ++ tiaoshi;
                if(S & (1 << (i - 1))) 
                    sum[S] += a[i], ++ num[S];
            }
        if(sum[ma - 1] != x * y) ans = 0;
        else ans = dfs(min(x, y), ma - 1);
        if(ans) printf("Case %d: Yes
", t);
        else printf("Case %d: No
", t);
    }
    return 0;
}