Poj2288 (islands and bridges) Pressure DP

Question link: http://poj.org/problem? Id = 2288

Each vertex has a weight VI and finds a Hamilton path. The weight of the path comes from three: 1. The sum of VI in the path 2 the sum of VI * VJ of all adjacent points to IJ 3 the sum of VI * VJ * VK of adjacent three consecutive I, j, and K (and the three points must form a triangle.

Solution: DP [st] [I] [J] indicates the maximum weight from J to I and the rest of the Set st is not taken. The path book can be computed by the way during the transfer. This question has been disgusting for a long time. The final reason is that you have made a serious mistake and read the wrong question, if you do not read VI * VJ * VK, you must ensure that ijk can form a triangle and check whether ik has edges by modifying the code. Of course, pay attention to the situations where the number of paths may exceed int and a specific point;

Code:

/******************************************************* @author:xiefubao*******************************************************/#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <vector>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <string.h>//freopen ("in.txt" , "r" , stdin);using namespace std;#define eps 1e-8#define zero(_) (abs(_)<=eps)const double pi=acos(-1.0);typedef long long LL;const int Max=14;const int INF=1e9+7;int num[Max];LL dp[1<<Max][Max][Max];LL cnt[1<<Max][Max][Max];bool rem[Max][Max];vector<int> vec[Max];int n,m;LL getdp(int st,int i,int j){    if(dp[st][i][j]!=-1)        return dp[st][i][j];    LL ans=-INF;    LL sum=0;    for(int l=0; l<vec[i].size(); l++)    {        int k=vec[i][l];        if((st&(1<<k))==0) continue;        if()        LL tool=getdp(st-(1<<k),k,i)+num[k]*num[i]+num[k]*num[i]*num[j];        if(tool==ans)            sum+=cnt[st-(1<<k)][k][i];        if(tool>ans)        {            sum=cnt[st-(1<<k)][k][i];            ans=tool;        }    }    cnt[st][i][j]=sum;    return dp[st][i][j]=ans;}int main(){    int t;    cin>>t;    while(t--)    {        memset(rem,0,sizeof rem);        scanf("%d%d",&n,&m);        int sum=0;        for(int i=0; i<n; i++)            scanf("%d",num+i),sum+=num[i],vec[i].clear();        for(int i=0; i<m; i++)        {            int a,b;            scanf("%d%d",&a,&b);            a--,b--;            if(a==b)                continue;            vec[a].push_back(b);            vec[b].push_back(a);            rem[a][b]=1;            rem[b][a]=1;        }        if(n==1)        {            printf("%d %d\n",num[0],1);            continue;        }        memset(dp,-1,sizeof dp);        memset(cnt,0,sizeof cnt);        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)            {                dp[0][i][j]=-INF;                if(!rem[i][j])continue;                dp[0][i][j]=0;                cnt[0][i][j]=1;            }        LL ans=-INF;        LL out=0;        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)            {                if(!rem[i][j])                    continue;                LL tool=getdp((1<<n)-(1<<i)-(1<<j)-1,i,j)+num[i]*num[j];                if(ans==tool)                    out+=cnt[(1<<n)-(1<<i)-(1<<j)-1][i][j];                if(ans<tool)                {                    ans=tool;                    out=cnt[(1<<n)-(1<<i)-(1<<j)-1][i][j];                }            }        if(out==0||ans<=0)            printf("0 0\n");        else            cout<<ans+sum<<" "<<out/2<<endl;    }    return 0;}/*6 151 1 1 1 1 11 21 31 41 51 62 32 42 52 63 43 53 64 54 65 6*/