hdu1074--Doing Homework（状压dp）

 2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3 

 2
Computer
Math
English
3
Computer
English
Math 

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. 

给出了n门作业，每科做也的提交时间和需要时间，每超过一天罚1分，问最少罚几分，要求最小字典序

状态表示，在n科中做了哪几科，二进制数中第i个为1表示做了第i科，dp[i]表示状态为i时的最少罚时，和已经用了的分数

状态转移方程从科目为j的转移到科目为j+1的

为保证字典序最小，所以要让dp[i]尽量有更小的i来得到。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
struct node
{
    char s[120] ;
    int d , c ;
} p[20];
int state[20][7000] , num[20] , c[20] , cnt ;
int dp1[20][7000] , dp2[20][7000] , step[20][7000] ; //dp[i][j][]写i科作业在状态j时，0代表超出的时间，1代表需要的时间。
int find1(int x,int i)
{
    //printf("x == %d i == %d\n", x, i) ;
    int low = 0 , mid , high = num[i]-1 ;
    while( low <= high )
    {
        mid = ( low + high ) ;
        //printf("mid--%d\n", mid) ;
        if( state[i][mid] == x )
            return mid ;
        else if( state[i][mid] > x )
            high = mid - 1 ;
        else
            low = mid + 1 ;
    }
}
int main()
{
    int t , n , i , j , k , l , x , y , temp ;
    scanf("%d", &t) ;
    while( t-- )
    {
        memset(num,0,sizeof(num)) ;
        memset(dp1,-1,sizeof(dp1)) ;
        memset(step,-1,sizeof(step)) ;
        memset(dp2,-1,sizeof(dp2));
        scanf("%d", &n) ;
        for(i = 1 ; i <= n ; i++)
            scanf("%s %d %d", p[i].s, &p[i].d, &p[i].c) ;
        x = 1<<n ;
        for(i = 0 ; i < x ; i++)
        {
            for(j = 0 , temp = 0 ; j < n ; j++)
                if( (1<<j) & i )
                    temp++ ;
            state[temp][ num[temp]++ ]=  i ;
        }
        /*
        for(i = 0 ; i <= n ; i++)
        {
            for(j = 0 ;  j < num[i] ; j++)
                printf("%d ", state[i][j]) ;
            printf("\n") ;
        }*/
        dp1[0][0] = dp2[0][0] = 0 ;
        for(i = 0 ; i < n ; i++)
        {
            for(j = 0 ; j < num[i] ; j++)
            {
                for(k = 1 ; k <= n ; k++)
                {
                    if( 1<<(k-1) & state[i][j] )
                        continue ;
                    l = find1(state[i][j]+(1<<(k-1)),i+1) ;
                    //printf("l == %d\n", l ) ;
                    if( dp1[i+1][l] == -1 || ( dp1[i+1][l] > dp1[i][j] + max(dp2[i][j]+p[k].c-p[k].d,0) ) )
                    {
                        dp1[i+1][l] = dp1[i][j] + max(dp2[i][j]+p[k].c-p[k].d,0) ;
                        dp2[i+1][l] = dp2[i][j]+p[k].c ;
                        step[i+1][l] = j ;
                    }
                }
            }
        }
        /*for(i = 0 ; i <= n ; i++)
        {
            for(j = 0 ; j < num[i] ; j++)
                printf("%d ", dp[i][j][0]) ;
            printf("\n") ;
        }*/
        cnt = 0 ;
        for(i = n , j = 0 ; i > 0 ; i--)
        {
            x = state[i][j] ;
            y = state[i-1][ step[i][j] ] ;
            //printf("x = %d y = %d\n", x, y) ;
            for(k = 1 ; k <= n ; k++)
            {
                if( x % 2 != y % 2 )
                    break ;
                x /= 2 ;
                y /= 2 ;
            }
            c[cnt++] = k ;
            j = step[i][j] ;
        }
        int min1 = dp1[n][0] ;
        printf("%d\n", min1) ;
        for(i = cnt - 1 ; i >= 0 ; i--)
            printf("%s\n", p[c[i]].s ) ;
    }
    return 0 ;
}

hdu1074--Doing Homework（状压dp）

原文：http://blog.csdn.net/winddreams/article/details/42586359