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UVA 11125 Arrange Some Marbles

2014-09-25 13:44 471 查看
dp[i][j][m]
[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\n",slove());
}
return 0;
}
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