hdu 4049 Tourism Planning 状态压缩DP_acm tourism planning-优快云博客

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/6843938

本文介绍了一个复杂的旅行计划问题，该问题旨在最大化参与者的整体满意度，同时考虑个人兴趣、团队奖金和旅行成本等因素。通过动态规划的方法，算法能够决定哪些人参加旅行、何时离开，以使总收益最大化。

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Problem Description

Several friends are planning to take tourism during the next holiday. They have selected some places to visit. They have decided which place to start their tourism and in which order to visit these places. However, anyone can leave halfway during the tourism and will never back to the tourism again if he or she is not interested in the following places. And anyone can choose not to attend the tourism if he or she is not interested in any of the places.
Each place they visited will cost every person certain amount of money. And each person has a positive value for each place, representing his or her interest in this place. To make things more complicated, if two friends visited a place together, they will get a non negative bonus because they enjoyed each other’s companion. If more than two friends visited a place together, the total bonus will be the sum of each pair of friends’ bonuses.
Your task is to decide which people should take the tourism and when each of them should leave so that the sum of the interest plus the sum of the bonuses minus the total costs is the largest. If you can’t find a plan that have a result larger than 0, just tell them to STAY HOME.

Input

There are several cases. Each case starts with a line containing two numbers N and M ( 1<=N<=10, 1<=M<=10). N is the number of friends and M is the number of places. The next line will contain M integers Pi (1<=i<=M) , 1<=Pi<=1000, representing how much it costs for one person to visit the ith place. Then N line follows, and each line contains M integers Vij (1<=i<=N, 1<=j<=M), 1<=Vij<=1000, representing how much the ith person is interested in the jth place. Then N line follows, and each line contains N integers Bij (1<=i<=N, 1<=j<=N), 0<=Bij<=1000, Bij=0 if i=j, Bij=Bji.
A case starting with 0 0 indicates the end of input and you needn’t give an output.

Output

For each case, if you can arrange a plan lead to a positive result, output the result in one line, otherwise, output STAY HOME in one line.

Sample Input

Sample Output

  
  
   
   5
41
STAY HOME

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=(1<<28);
int n,m;
int dp[12][1<<11];//dp[i][j]表示前i个地方,去的人的状态是j的最大兴趣值
int cost[12];
int interest[12][12];
int bonus[12][12];
int geti(int x,int state)//第x个地方给state状态的人的加入带来的效益
{
int cnt=0;
for(int i=0;i<n;i++)
{
if(state&(1<<i))
{
cnt-=cost[x];
cnt+=interest[i][x];
for(int j=i+1;j<n;j++)//important i+1 防止重复计算
{
if(state&(1<<j))
{
cnt+=bonus[i][j];
}
}
}
}
return cnt;
}
int main()
{
while(scanf("%d%d",&n,&m)==2&&n)
{
for(int i=0;i<m;i++)scanf("%d",&cost[i]);
for(int i=0;i<n;i++) for(int j=0;j<m;j++) scanf("%d",&interest[i][j]);
for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%d",&bonus[i][j]);
for(int i=0;i<(1<<n);i++)
{
dp[0][i]=geti(0,i);
}
for(int i=1;i<m;i++)
{
for(int j=0;j<(1<<n);j++)
{
int cnt=geti(i,j);//每两个朋友访问同一个地方,就要加上一次bonus
dp[i][j]=dp[i-1][j]+cnt;
for(int k=j+1;k<(1<<n);k++)
{
if((k&j)==j)//旅游过程中可以离开,所以i-1要比i的人多,且i的人i-1都有
{
dp[i][j]=max(dp[i][j],dp[i-1][k]+cnt);
}
}
}
}
int ans=0;
for(int i=0;i<(1<<n);i++) ans=max(ans,dp[m-1][i]);
if(ans) printf("%d\n",ans);
else printf("STAY HOME\n");
}
return 0;
}