北京1009 hdu4049 Tourism Planning DP初始化

 

Tourism Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 395    Accepted Submission(s): 168

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

2 1101550 55 03 230 5024 4840 7035 200 4 14 0 51 5 02 2100 10050 5050 500 2020 00 0

 

Sample Output

541STAY HOME

 

Source

The 36th ACM/ICPC Asia Regional Beijing Site —— Online Contest

 

Recommend

lcy

 

/*犯的错误：开始没有初始化，后面觉得某个人一直走到最后会是正的，就一定要走到最后*/#include<string.h>#include<iostream>#include <stdio.h>#include <memory.h>using namespace std;//typedef long long LL;typedef __int64 LL;const int maxn=1<<11;LL A[11][11],B[11],g[11][11],dp[12][maxn],w[12][maxn],ww[maxn][maxn],m,n;LL Cal(){LL i1,j1,k1,ii,kk,i,j,k,a[15];for(i=0;i<m;i++)for(j=0;j<1<<n;j++)dp[i][j]=-100000000;memset(w,0,sizeof(w));for(j=0;j<1<<n;j++){i=j,ii=0,k=0;while(i){if(i&1)a[ii++]=k;k++;i=i>>1;}for(kk=0;kk<m;kk++){for(i=0;i<ii;i++)w[kk][j]+=A[a[i]][kk]-B[kk];for(i=0;i<ii;i++)for(k=i+1;k<ii;k++)w[kk][j]+=g[a[i]][a[k]];}}for(j=0;j<1<<n;j++)dp[0][j]=w[0][j];for(i=0;i<1<<n;i++){for(k=1,j=0;j<1<<n;j++)if((j|i)==i)ww[i][k++]=j;ww[i][0]=k;}for(i1=1;i1<m;i1++)//在i1点，在i1-1点剩的人的二进制表示为k1{for(j1=0;j1<1<<n;j1++){for(k1=1;k1<ww[j1][0];k1++){ii=ww[j1][k1];if(dp[i1][ii]<dp[i1-1][j1]+w[i1][ii])dp[i1][ii]=dp[i1-1][j1]+w[i1][ii];}}}for(kk=0,i=0;i<1<<n;i++)if(dp[m-1][i]>kk) kk=dp[m-1][i];return kk;}void Init(){    LL i,j;    for(i=0;i<m;i++)//每个人进所需的钱        scanf("%I64d",&B[i]);    for(i=0;i<n;i++)//i人到j地方的兴趣        for(j=0;j<m;j++)scanf("%I64d",&A[i][j]);    for(i=0;i<n;i++)//i与j同时进再加的兴趣        for(j=0;j<n;j++)            scanf("%I64d",&g[i][j]);}int main(){LL ii;while(scanf("%I64d%I64d",&n,&m)!=EOF){if(n==0&&m==0) break;Init();ii=Cal();if(ii<=0) printf("STAY HOME\n");else printf("%I64d\n",ii);}return 0;}