Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test, 1 day for 1 point. And as you know, doing homework always takes a long time. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject’s name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject’s homework).
Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.
Output
For each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.
Sample Input
2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3
Sample Output
2
Computer
Math
English
3
Computer
English
Math
Hint
In the second test case, both Computer-> English-> Math and Computer-> Math-> English leads to reduce 3 points, but the
word “English” appears earlier than the word “Math”, so we choose the first order . That is so-called alphabet order.
Idea: Because n = 15, it is not very large, so we can enumerate all the states, continuously update the minimum value of each state, and finally output dp [(1 << n) -1]. See the code specifically:
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxx=16;
const int maxn=(1<<maxx);
struct node{
string s;
int ed,vl;
}p[maxx];
int dp[maxn],tme[maxn],pre[maxn],bk[maxn];
int n;
inline void init()
{
memset(dp,0,sizeof(dp));
memset(pre,-1,sizeof(pre));
memset(tme,0,sizeof(tme));
}
inline void dfs(int x)
{
if(x==0) return ;
dfs(pre[x]);
cout<<p[bk[x]].s<<endl;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++) cin>>p[i].s>>p[i].ed>>p[i].vl;
init();
int All=1<<n;All-=1;
int tmp,sum;
for(int i=1;i<=All;i++)//枚举状态，这里枚举的就是每一种可能发生的情况
{
dp[i]=inf;
for(int j=n;j>=1;j--)//正序的话，最小值计算的是没有问题的，但是无法按照题目要求的顺序输出
{
tmp=1<<(j-1);
if(!(i&tmp)) continue;
sum=tme[i-tmp]+p[j].vl-p[j].ed;//i-tmp，例如i=001010,tmp=001000,i-tmp=000010.就代表第五个作业先做，第三个作业再做。
if(sum<0) sum=0;
if(sum+dp[i-tmp]<dp[i])
{
dp[i]=dp[i-tmp]+sum;
tme[i]=tme[i-tmp]+p[j].vl;
pre[i]=i-tmp;
bk[i]=j;
}
}
}
printf("%d\n",dp[All]);
dfs(All);
}
return 0;
}
To be honest, I didn't expect it to be a pressure at first. .
Come on a hard, ( o ) / ~
