HOJ 2317 Pimp My Ride(状态压缩DP)

Pimp My Ride

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Source : TUD 2005

Time limit : 3 sec Memory limit : 64 M

Submitted : 63, Accepted : 34

Today, there are quite a few cars, motorcycles, trucks and other vehicles out there on the streets that would seriously need some refurbishment. You have taken on this job, ripping off a few dollars from a major TV station along the way.

Of course, there’s a lot of work to do, and you have decided that it’s getting too much. Therefore you want to have the various jobs like painting, interior decoration and so on done by garages. Unfortunately, those garages are very specialized, so you need different garages for different jobs. More so, they tend to charge you the more the better the overall appearance of the car is. That is, a painter might charge more for a car whose interior is all leather. As those “surcharges” depend on what job is done and which jobs have been done before, you are currently trying to save money by finding an optimal order for those jobs.

Problem

Individual jobs are numbered 1 through n. Given the base price p for each job and a surcharge s (in US)foreverypairofjobs(i,j)withi!=j,meaningthatyouhavetopayadditional) for every pair of jobs (i, j) with i != j, meaning that you have to pay additional s for job i, if and only if job j was completed before, you are to compute the minimum total costs needed to finish all jobs.

Input

The first line contains the number of scenarios. For each scenario, an integer number of jobs n, 1 <= n <= 14, is given. Then follow n lines, each containing exactly n integers. The i-th line contains the surcharges that have to be paid in garage number i for the i-th job and the base price for job i. More precisely, on the i-th line, the i-th integer is the base price for job i and the j-th integer (j != i) is the surcharge for job i that applies if job j has been done before. The prices will be non-negative integers smaller than or equal to 100000.

Output

The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line: “You have officially been pimped for only p” with p being the minimum total price. Terminate the output for the scenario with a blank line.
Sample Input
2
2
10 10
9000 10
3
14 23 0
0 14 0
1000 9500 14
Sample Output
Scenario #1:
You have officially been pimped for onlyp” with p being the minimum total price. Terminate the output for the scenario with a blank line.
Sample Input
2
2
10 10
9000 10
3
14 23 0
0 14 0
1000 9500 14
Sample Output
Scenario #1:
You have officially been pimped for only 30

Scenario #2:

You have officially been pimped for only $42

注意题目：每次做一个job都要把之前做过的所有job额外的费用加上

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 100000000
int dp[1<<15][15];
int a[15][15];
int ans;
int n;
int main()
{
int t;
scanf("%d",&t);
int cas=0;
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
int state=(1<<n)-1;
for(int i=0;i<=state;i++)
for(int j=0;j<n;j++)
dp[i][j]=MAX;
for(int i=0;i<n;i++)
dp[1<<i][i]=a[i][i];
for(int i=1;i<=state;i++)
{
for(int j=0;j<n;j++)
{
if((1<<j)&i)
{
int num=a[j][j];
for(int k=0;k<n;k++)
{
if(k!=j&&((1<<k)&i))
num+=a[j][k];
}
for(int p=0;p<n;p++)
{
if(p!=j&&((1<<p)&i))
dp[i][j]=min(dp[i][j],num+dp[i-(1<<j)][p]);
}
}

}
}
ans=MAX;
for(int i=0;i<n;i++)
ans=min(ans,dp[state][i]);
printf("Scenario #%d:\n",++cas);
printf("You have officially been pimped for only $%d\n\n",ans);

}
return 0;
}