ACM: DP训练好题 动态规划题 poj 1…

CowCycling

 

Description

The cow bicyclingteam consists of N (1 <= N <= 20)cyclists. They wish to determine a race strategy which will get oneof them across the finish line as fast as possible.

Like everyone else, cows race bicycles in packs because that's themost efficient way to beat the wind. While travelling at xlaps/minute (x is always an integer), the head of the pack expendsx*x energy/minute while the rest of pack drafts behind him usingonly x energy/minute. Switching leaders requires no time though canonly happen after an integer number of minutes. Of course, cows candrop out of the race at any time.

The cows have entered a race D (1 <= D<= 100) laps long. Each cow has the same initialenergy, E (1 <= E <= 100).

What is the fastest possible finishing time? Only one cow has tocross the line. The finish time is an integer. Overshooting theline during some minute is no different than barely reaching it atthe beginning of the next minute (though the cow must have theenergy left to cycle the entire minute). N, D, and E areintegers.

Input

A single line withthree integers: N, E, and D

Output

A single line withthe integer that is the fastest possible finishing time for thefastest possible cow. Output 0 if the cows are not strong enough tofinish the race.

Sample Input

3 30 20

Sample Output

7

Hint

[as shown in this chart:

                           leader E

              pack  total used this

time  leader speed  dist   minute

 1     1     5      5     25

 2     1     2      7      4

 3     2*    4     11     16

 4     2     2     13      4

 5     3*    3     16      9

 6     3     2     18      4

 7     3     2     20      4

* = leader switch

 

题意: 现在又N头奶牛, 每头奶牛的能量是E, 现在要奶牛们只要有一头完成跑D圈就完成任务.

      但是每次都要由一头奶牛领跑,没有能量的奶牛可以退场或继续跟跑, 但是带头的奶牛

     体力消耗是: x*x laps/min, 跟跑的是x laps/min. 现在要你计算出最短时间内完成任务.

 

解题思路:

         1.题意应该很明确, 前n-1头奶牛都领跑过, 第n头奶牛冲线.

        2. 状态: dp[i][j][k]: 表示第i头奶牛领跑, 跑了j圈, 第i头奶牛体力消耗了k的最小时间.

        3. 状态转移方程:

               (1). dp[i][j+p][k+p^2] = min( dp[i][j+p][k+p^2] ,dp[i][j][k]+1 );

               (2). dp[i+1][j][j] = min( dp[i+1][j][j] ,dp[i][j][k] ) //换一头奶牛领跑不耗时.

 

代码:

#include<cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int INF = (1<<29);

int n, e, d;
int dp[25][105][105];

inline int min(int a,int b)
{
 return a < b ? a : b;
}

int main()
{
 int i, j, k;
// freopen("input.txt","r",stdin);
 while(scanf("%d %d%d",&n,&e,&d) !=EOF)
 {
  for(i = 0; i <=n; ++i)
   for(j = 0; j<= d; ++j)
    for(k= 0; k <= e; ++k)
     dp[i][j][k]= INF;
  dp[1][0][0] = 0;
  for(i = 1; i <=n; ++i)
  {
   for(j = 0; j<= d; ++j)
   {
    for(k= 0; k <= e; ++k)
    {
     if(dp[i][j][k]== INF) continue; //无意义
     for(intp = 1; p+j <= d &&p*p+k <= e; ++p)
      dp[i][j+p][k+p*p]= min(dp[i][j][k]+1,dp[i][j+p][k+p*p]);
     dp[i+1][j][j]= min(dp[i+1][j][j], dp[i][j][k]);
    }
   }
  }

  int result =INF;
  for(i = 0; i <=e; ++i)
   result =min(result,dp[n][d][i]);
  printf("%d\n",result);//当无法完成任务的时候dp[n][d][0] = 0;
 }
 return 0;
}