UVA 11125 Arrange Some Marbles

dp[i][j][m][n][s]表示最初选择j个i号颜色大理石。当前选择n个m号颜色大理石。剩余大理石状态（8进制数状压表示）最开始没看出状压。。sad

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
int base_8[]={1 ,8 ,64 ,512};
int cnt[4],dp[4][4][4][4][5000];
int N;
int calcu(int hc, int hl, int pc, int pl, int sta)
{
    if (dp[hc][hl][pc][pl][sta]!=-1) return dp[hc][hl][pc][pl][sta];
    dp[hc][hl][pc][pl][sta]=0;
    if (sta == 0)
    {
        if (hc != pc && hl != pl) return dp[hc][hl][pc][pl][sta] = 1;
        else return dp[hc][hl][pc][pl][sta] = 0;
    }
    for (int i = 0; i < N; i++)
        for (int j = 1; j <= 3 && j <= cnt[i]; j++)
    {
        if (i != pc && j != pl)
        {
            cnt[i] -= j;
            dp[hc][hl][pc][pl][sta] += calcu(hc, hl, i, j, sta - base_8[i] * j);
            cnt[i] += j;
        }
    }
    return dp[hc][hl][pc][pl][sta];
}
int slove()
{
    int state=0,ans=0;
    for (int i = N - 1; i >= 0; i--) state = state * 8 + cnt[i];
    for (int i = 0; i < N; i++)
        for (int j = 1; j <= 3 && j <= cnt[i]; j++)
    ans += calcu(i, j, i, j,state - j * base_8[i]);
    return ans;
}
int main()
{
    int T;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&N);
        for (int i = 0; i < N; i++) scanf("%d",&cnt[i]);
        if (cnt[0] == 0 && cnt[1] == 0 && cnt[2] == 0 && cnt[3] == 0) puts("1");
        else  printf("%d
",slove());
    }
    return 0;
}