poj2288(状压dp)

这题如果没想限制有三个成环的时候是可以一维转压过的，但是这个条件。。。。所以要设计这样的状态dp[s][i][j]在状态s时现在点是i上个点是j的最大值。

枚举上两个点，然后枚举这个点。

#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<string>
#include<vector>
#include<queue>
#include<list>
using namespace std;
#define ep 1e-9
#define oo 0x3f3f3f3f
typedef __int64 lld;
lld dp[1 << 14][15][15];
lld num[1 << 14][15][15];
int map[15][15];
lld val[15];
int n, m;

int main()
{
	int T, n, m, u, v;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d %d", &n, &m);
		for (int i = 1; i <= n; i++)
			scanf("%I64d", &val[i]);
		memset(map, 0, sizeof map);
		for (int i = 1; i <= m; i++)
		{
			scanf("%d %d", &u, &v);
			map[u][v] = map[v][u] = 1;
		}
		if (n == 1)
		{
			printf("%I64d 1\n", val[1]);
			continue;
		}

		memset(dp, -1, sizeof dp);
		memset(num, 0, sizeof num);
		int all = 1 << n;
		for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			if (i == j || !map[i][j]) continue;
			int s1 = 1 << (i - 1);
			int s2 = 1 << (j - 1);
			dp[s1 | s2][i][j] = val[i] + val[j] + val[i] * val[j];
			num[s1 | s2][i][j] = 1;
		}
		for (int i = 0; i < all; i++)
		for (int j = 1; j <= n; j++)
		{
			if (!(i&(1 << (j - 1)))) continue;
			for (int k = 1; k <= n; k++)
			{
				if (k == j || !(i&(1 << (k - 1))) || !map[j][k]) continue;
				if (dp[i][j][k] == -1) continue;

				for (int t = 1; t <= n; t++)
				{
					if (t == k || t == j) continue;
					if ((i&(1 << (t - 1)))) continue;
					if (!map[k][t]) continue;
					int st = i|(1 << (t - 1));

					lld sum = 0;
					sum += val[t] + val[t] * val[k];
					if (map[j][t]) sum += val[j] * val[k] * val[t];

					if (dp[i][j][k] + sum > dp[st][k][t])
					{
						dp[st][k][t] = dp[i][j][k] + sum;
						num[st][k][t] = num[i][j][k];
					}
					else if (dp[i][j][k] + sum == dp[st][k][t])
						num[st][k][t] += num[i][j][k];
				}
			}
		}
		lld max = 0, sum = 0;
		for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			if (max < dp[all - 1][i][j])
			{
				max = dp[all - 1][i][j];
				sum = num[all - 1][i][j];
			}
			else if (max == dp[all - 1][i][j])
				sum += num[all - 1][i][j];
		}
		printf("%I64d %I64d\n", max, sum / 2);
	}
	return 0;
}