Doing Homework
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6037 Accepted Submission(s): 2566
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.
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define M 100000
using namespace std;
struct statue
{
int reduce,pre,cost;
}dp[M];
struct subject
{
char name[111];
int cost,deadline;
}sub[20];
void output(int i)
{
if(dp[i].pre==-1) return;
output(dp[i].pre);
int a=i&~dp[i].pre,bit=0;
while(a!=0)
{
bit++;
a/=2;
}
cout<<sub[bit-1].name<<endl;
}
int main(int argc, char *argv[])
{
int t,n,i,j;
cin>>t;
while(t--)
{
cin>>n;
for(i=0;i<n;i++)
{
scanf("\n%s%d%d",sub[i].name,&sub[i].deadline,&sub[i].cost);
}
for(i=0;i<M;i++) dp[i].reduce=9999999;
const int N=(int)(pow(2.0,n)-1);
dp[0].pre=-1;dp[0].reduce=0;dp[0].cost=0;
for(i=1;i<=N;i++)
{
int bit=0,ii=i;
while(ii!=0)
{
bit++;
ii/=2;
}
for(j=0;j<bit;j++)
{
if(i&(1<<j))
{
dp[i].cost=dp[i&~(1<<j)].cost+sub[j].cost;
int r,jj;
if(dp[i].cost>sub[j].deadline)
r=dp[i&~(1<<j)].reduce+dp[i].cost-sub[j].deadline;
else r=dp[i&~(1<<j)].reduce;
if(r<dp[i].reduce)
{
dp[i].reduce=r;
dp[i].pre=i&~(1<<j);
jj=j;
}
else if(r==dp[i].reduce&&strcmp(sub[j].name,sub[jj].name)==1)
{
dp[i].reduce=r;
dp[i].pre=i&~(1<<j);
jj=j;
}
}
}
}
cout<<dp[N].reduce<<endl;
output(N);
/*for(i=0;i<=N;i++)
{
cout<<dp[i].pre<<" "<<dp[i].cost<<" "<<dp[i].reduce<<endl;
}*/
}
return 0;
}