HDU 1074 Doing Homework 状态压缩 + 搜索

题意：

就是有0-15个工作，每个工作有完成日期限制，还有完成这个工作所需要时间，求完成所有工作花时最少，并且输出这些顺序。。。

因为工作很小，最多也就1<<15种方案，BFS枚举所有方案即可。

/*
鐘舵€佸帇缂?+ 鎼滅储
*/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;

struct node
{
 char name[110];
 int end;
 int day; 
}Work[20];

int N;
struct Qnode
{
  int cost; //鏈€灏戞崯澶卞垎鏁?
  int v;//鍓嶇姸鎬佸€?
  int work; //宸ヤ綔澶╂暟
  char s[20];
}dp[40000];

void bfs(  )
{
  queue<Qnode>q;
  int state, cost, day;
  Qnode xx, yy;
  xx.cost = xx.v = xx.work = 0;
  xx.s[0] = '\0';
  dp[0] = xx;
  q.push(xx);
  while( !q.empty() )
  {
    xx = q.front( );
    state = xx.v;
    cost = xx.cost;
    day = xx.work;
    q.pop( );
    for( int i = 0; i < N; i++)
    {
       if( ((1<<i)&state) == 0 ) //濡傛灉璇ヨ鏈闂?
       {
          yy.v = (1<<i)|state;
          yy.work = day + Work[i].day;
          yy.cost = cost;
          if( yy.work > Work[i].end )
              yy.cost += yy.work - Work[i].end;
          if( dp[yy.v].cost == -1 || dp[yy.v].cost > yy.cost )
          {
               strcpy(yy.s, xx.s);
               int len = strlen(yy.s);
               yy.s[len] = i + '0';
               yy.s[len+1] = '\0';
               dp[yy.v] = yy;
               q.push(yy);
          }
       }
    }
  }
}

int main( )
{
  int T;
  scanf("%d",&T);
  while(T--)
  {
    memset(dp, 0, sizeof(dp));
    scanf("%d",&N);
    for( int i = 0; i < N; i++)
    {
      scanf("%s%d%d",Work[i].name, &Work[i].end, &Work[i].day);
    }
    for( int i = 0; i < (1<<N); i++)
       dp[i].cost = -1;
    bfs( );
    printf("%d\n", dp[(1<<N)-1].cost);
    for( int i = 0; i < N; i++)
    printf("%s\n", Work[dp[(1<<N)-1].s[i] - '0'].name);
  }
  return 0;
}