HDU 1074
慢慢写了半个多小时的dp,思路很清楚，二进制保存状态，然后如果这个状态为0，就添上1转移，并且记录路径，做之前一定要想好要记录哪些东西，由于一个状态不管cost是啥，总之是一样的。因为不管什么状态，加到这一步，总天数都是一样的。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
int vis[1<<15+5];
struct Work
{
char name[110];
int deadline;
int cost;
}date[20];
struct Go
{
int pre;
int reduce;
int cost;
}dp[1<<15+5];
void out(int temp)
{
int tempnext=dp[temp].pre;
if (tempnext!=-1)
{
int a=(int)(log(temp-tempnext)/log(2));
out(tempnext);
printf ("%s\n",date[a].name);
}
else return;
}
int main()
{
int T,n;
scanf ("%d",&T);
while(T--)
{
scanf ("%d",&n);
memset(vis,0,sizeof(vis));
memset(dp,0,sizeof(dp));
for (int i=0;i<n;i++)
{
scanf ("%s%d%d",date[i].name,&date[i].deadline,&date[i].cost);
}
int maxdp=(1<<n)-1;
dp[0].pre=-1;
dp[0].reduce=0;
dp[0].cost=0;
for (int i=0;i<maxdp;i++)
{
for (int j=0;j<n;j++)
{
int cur=1<<j;
if ((i&cur)==0)
{
int curnext=cur|i;
int day=dp[i].cost+date[j].cost;
dp[curnext].cost=day;
int rreduce=day-date[j].deadline;
if (rreduce<=0) rreduce=0;
rreduce+=dp[i].reduce;
if (vis[curnext])
{
if (rreduce<dp[curnext].reduce)
{
dp[curnext].reduce=rreduce;
dp[curnext].pre=i;
}
}
else
{
vis[curnext]=1;
dp[curnext].pre=i;
dp[curnext].reduce=rreduce;
}
}
}
}
printf ("%d\n",dp[maxdp].reduce);
out(maxdp);
}
return 0;
}