State compression ----- hdu1074 doing homework

Hdu1074 doing homework

　　

Question: I gave n homework assignments, and then gave each homework a completion time and spend. If the completion time exceeds the time limit, I would deduct the score, then let you find an optimal solution to minimize the number of points deducted. When there are multiple solutions, the one with the smallest Lexicographic Order is output, the meaning of the question has already been said that the names of homework are input in Lexicographic Order from small to large, so it is much better to deal with it.

Analysis: the Key to this question is how to record the path. For details, refer to the code.

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5  6 using namespace std; 7 #define MAX 1<<15+1 8 int N; 9 struct Node{10 char name[108];11 int deadline;12 int cost;13 }homework[16];14 struct START15 {16     int time;17     int pre;18     int score;19     int now;20 } dp[MAX];21 int record[16];22 23 void DP()24 {25     int Final=(1<<N)-1,punish,previous;26     dp[0].id=dp[0].now=dp[0].pre=dp[0].score=dp[0].time=0;27     for(int i=1;i<=Final;i++)28     {29         dp[i].score=10000000;30         for(int j=0;j<N;j++)31             if(i&(1<<j))32         {33              previous=(i-(1<<j));34             if(dp[previous].time+homework[j].cost>homework[j].deadline)35                 punish=dp[previous].time+homework[j].cost-homework[j].deadline;36             else punish=0;37             if(dp[i].score>=dp[previous].score+punish)38             {39                 dp[i].score=dp[previous].score+punish;40                 dp[i].time =dp[previous].time+homework[j].cost;41                 dp[i].pre=previous;42                 dp[i].now=j;43             }44         }45     }46     printf("%d\n",dp[Final].score);47     int cal=Final,num=0;48     while(cal)49     {50         record[num++]=dp[cal].now;51         cal=dp[cal].pre;52     }53     for(int i=num-1;i>=0;i--)54     puts(homework[record[i]].name);55 56 }57 int main()58 {59    int T;60    scanf("%d",&T);61    while(T--)62    {63        scanf("%d",&N);64        for(int i=0;i<N;i++)scanf("%s%d%d",homework[i].name,&homework[i].deadline,&homework[i].cost);65        DP();66    }67     return 0;68 }

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It was frustrating to do this. In DP (), J was written as J + 1, which was not visible for a long time. After debugging for nearly one hour, I found it.

1 void dp () 2 {3 int final = (1 <n)-1, punish, previous; 4 DP [0]. id = DP [0]. now = DP [0]. pre = DP [0]. score = DP [0]. time = 0; 5 for (INT I = 1; I <= final; I ++) // The status ranges from 0 to 6 {7 DP [I]. score = 10000000; // first initialized to a relatively large value of 8 for (Int J = 0; j <n; j ++) to find the minimum value) // question No. 9 if (I & (1 <j) // question J in status I, why is this written because every possibility is enumerated at 10 // when the status is full, that is, the status is converted from previous to J11 {12 previous = (I-(1 <j )); // only one job is different from previous to I. // if the total time from previous to another question exceeds the deadline of job J, record points 14 // otherwise 0 15 if (DP [previous]. time + homework [J]. cost> homework [J]. deadline) 16 punish = DP [previous]. time + homework [J]. cost-Homework [J]. deadline; 17 else punish = 0; 18 // the minimum value of 19 if (DP [I] When the status is I. score> = DP [previous]. score + punish) 20 {21 DP [I]. score = DP [previous]. score + punish; 22 DP [I]. time = DP [previous]. time + homework [J]. cost; 23 DP [I]. pre = previous; 24 DP [I]. now = J; 25} 26} 27} 28 printf ("% d \ n", DP [Final]. score); 29 // The display path now indicates that status I is formed by a question numbered now, and previous is the previous state of I, 30 int Cal = final, num = 0; 31 While (CAL) 32 {33 record [num ++] = DP [Cal]. now; 34 Cal = DP [Cal]. pre; 35} 36 for (INT I = num-1; I> = 0; I --) 37 puts (homework [Record [I]. name); 38 39}

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