POJ 2288 Islands and Bridges（状压dp）

Language: Default Islands and Bridges Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 9312   Accepted: 2424 Description Given a map of islands and bridges that connect these islands, a Hamilton path, as we all know, is a path along the bridges such that it visits each island exactly once. On our map, there is also a positive integer value associated with each island. We call a Hamilton path the best triangular Hamilton path if it maximizes the value described below.  Suppose there are n islands. The value of a Hamilton path C1C2...Cn is calculated as the sum of three parts. Let Vi be the value for the island Ci. As the first part, we sum over all the Vi values for each island in the path. For the second part, for each edge CiC i+1 in the path, we add the product Vi*V i+1. And for the third part, whenever three consecutive islands CiC i+1C i+2 in the path forms a triangle in the map, i.e. there is a bridge between Ci and C i+2, we add the product Vi*V i+1*V i+2.  Most likely but not necessarily, the best triangular Hamilton path you are going to find contains many triangles. It is quite possible that there might be more than one best triangular Hamilton paths; your second task is to find the number of such paths.  Input The input file starts with a number q (q<=20) on the first line, which is the number of test cases. Each test case starts with a line with two integers n and m, which are the number of islands and the number of bridges in the map, respectively. The next line contains n positive integers, the i-th number being the Vi value of island i. Each value is no more than 100. The following m lines are in the form x y, which indicates there is a (two way) bridge between island x and island y. Islands are numbered from 1 to n. You may assume there will be no more than 13 islands.  Output For each test case, output a line with two numbers, separated by a space. The first number is the maximum value of a best triangular Hamilton path; the second number should be the number of different best triangular Hamilton paths. If the test case does not contain a Hamilton path, the output must be `0 0'.  Note: A path may be written down in the reversed order. We still think it is the same path. Sample Input 2 3 3 2 2 2 1 2 2 3 3 1 4 6 1 2 3 4 1 2 1 3 1 4 2 3 2 4 3 4 Sample Output 22 3 69 1 Source Shanghai 2004

题意： n个点，要求全部走一遍价值最大，价值的计算是 走过点的值相加  加上 每相邻的两个点的价值相乘，如果相邻的3个点组成三角形还要加上这三个值的乘积

思路 ：状压dp，我的解法是想初始化状态，还有看看当前的状态可以更新那些状态

// poj 2288

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 1<<14

ll dp[N][14][14];       // dp[s][i][j] 代表s状态中 i 到达 j 的价值
ll road[N][14][14];     // ........     .....                 路线
ll a[14][14],va[14];
ll n,m;

void solve()
{
	int i,j,k,cur;
	mem(dp,-1);
	mem(road,0);

	fre(i,1,n+1)
	 fre(j,1,n+1)
	  if(a[i][j]&&i!=j)
	  {
	  	 cur=1<<(i-1);
	  	 cur|=1<<(j-1);
	  	 dp[cur][i][j]=va[i]+va[j]+va[i]*va[j];  // i->j  初始化
	  	 road[cur][i][j]=1;
	  }

    ll len=1<<n;
    ll to;

    fre(cur,0,len)
     fre(i,1,n+1)
      {
      	 if(!(cur&(1<<(i-1)))) continue;   //没有经过 i

      	 fre(j,1,n+1)
		{
      	     if(!(cur&(1<<(j-1)))) continue;    //没有经过j

      	     if(!a[i][j]||i==j) continue;      //i j 没有路

      	     if(dp[cur][i][j]==-1) continue;     
      	              //此状态没有被到达过(因为用这个状态更新别的状态)

      	     fre(to,1,n+1)
      	     {
      	     	 if(a[j][to]==0) continue;   //枚举下一个到的点

      	     	 if(i==to||j==to) continue;

                 if(cur&(1<<(to-1))) continue;   //该状态没有to点

      	     	 int temp=dp[cur][i][j]+va[to]+va[j]*va[to];

      	     	 if(a[i][to]) temp+=va[i]*va[j]*va[to];   //形成三角形  

      	     	 int next=cur|(1<<(to-1));

                // if(next==len-1) bug;

      	     	 if(dp[next][j][to]<temp)
				 {
				 	 dp[next][j][to]=temp;
				 	 road[next][j][to]=road[cur][i][j];
				 }
                 else
					if(dp[next][j][to]==temp)
				   {
				 	  road[next][j][to]+=road[cur][i][j];
				   }
      	     }

		}
	}

	ll ans=0,ma=-1;
	len--;

	fre(i,1,n+1)
	 fre(j,1,n+1)
	 {
	 	if(i==j||a[i][j]==0) continue;
	 	if(dp[len][i][j]==-1) continue;

	 	if(dp[len][i][j]>ma)
		{
			ans=road[len][i][j];
			ma=dp[len][i][j];
		}
        else
			if(ma==dp[len][i][j])
			  ans+=road[len][i][j];
	 }


	if(ma==-1)
		printf("0 0\n");
	else
		printf("%I64d %I64d\n",ma,ans/2);


}

int main()
{
	int i,j,t;
	sf(t);

	while(t--)
	{

		scanf("%I64d%I64d",&n,&m);

		fre(i,1,n+1)
		  {
		  	scanf("%I64d",&va[i]);
		  }
		mem(a,0);
		ll s,e;

		while(m--)
		{
			scanf("%I64d%I64d",&s,&e);
			a[s][e]=a[e][s]=1;
		}


		  if(n==1)
		 {
			pf("%I64d 1\n",va[1]);
			continue;
		 }

	  solve();

	}
    return 0;

}