POJ 2279

线性DP

本题的正解是杨氏矩阵与钩子定理

但是这道题用DP的思想非常好



但是这样会MLE...

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
unsigned int dp[31][31][31][31][31], n, num[6];
int main() {
	while(1) {
		cin >> n;
		if(!n) break;
		memset(num, 0, sizeof(num));
		memset(dp, 0, sizeof(dp));
		dp[0][0][0][0][0] = 1;
		for(int i = 1; i <= n; i++) cin >> num[i];
		for(int i = 0;i <= num[1]; i++) {
			for(int j = 0; j <= num[2] && j <= i; j++) {
				for(int k = 0; k <= num[3] && k <= j; k++) {
					for(int l = 0; l <= num[4] && l <= k; l++) {
						for(int m = 0; m <= num[5] && m <= l; m++) {
							if(i + 1 <= num[1]) dp[i + 1][j][k][l][m] += dp[i][j][k][l][m];
							if(j + 1 <= num[2] && j + 1 <= i) dp[i][j + 1][k][l][m] += dp[i][j][k][l][m];
							if(k + 1 <= num[3] && k + 1 <= j) dp[i][j][k + 1][l][m] += dp[i][j][k][l][m];
							if(l + 1 <= num[4] && l + 1 <= k) dp[i][j][k][l + 1][m] += dp[i][j][k][l][m];
							if(m + 1 <= num[5] && m + 1 <= l) dp[i][j][k][l][m + 1] += dp[i][j][k][l][m];
						}
					}
				}
			}
		}
		cout << dp[num[1]][num[2]][num[3]][num[4]][num[5]] << endl;
	}
	return 0;
}