POJ 1946 Cow Cycling

Cow Cycling

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2516		Accepted: 1396

Description

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

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

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

What is the fastest possible finishing time?

Only one cow has to cross the line. The finish time is an integer. Overshooting the line during some minute is no different than barely reaching it at the beginning of the next minute (though the cow must have the
energy left to cycle the entire minute). N, D, and E are integers.

Input

A single line with three integers: N, E, and D 

Output

A single line with the integer that is the fastest possible finishing time for the fastest possible cow. Output 0 if the cows are not strong enough to finish 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

Source

USACO 2002 February

    我发现近期理解题意，真是该死，我理解成了最后仅仅能有一头牛通过，这样其它的牛除最后一头牛外最后的能量必须是负值。老是WA后。发现思路没错，就怀疑理解错题意。果然他要求的是仅仅要一头牛通过即可，求这头牛的最快时间，其它的牛不做要求。改了改初始话，AC了

题意：给定N头牛，每头牛的能量为E，距离为D，跑的时候假设是速度为X，则打头的牛的能量消耗为X*X/minute 其它的牛的能量消耗为X/minute 求当中一头牛到达终点时候，所用时间最少

思路：预处理出dp1[sta][end][e]:表示在能量为e的时候要跑的长度为sta，跑了end的时候用的时间最少

然后dfs+记忆化DP，枚举每头牛跑的长度。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#define N 110
#define INF 0x7ffffff
using namespace std;
int dp1[N][N][N];
int dp[N][N][N];
bool ch[N][N][N];
int main()
{
    //freopen("data.txt","r",stdin);
    int dfs(int n,int m,int k);
    int n,m,k;
    while(scanf("%d %d %d",&n,&m,&k)!=EOF)
    {
        for(int i=0;i<=k;i++)
        {
            for(int j=0;j<=i;j++)
            {
                for(int z = 0;z<=m;z++)
                {
                    dp1[i][j][z]=INF;
                }
            }
        }
        for(int z= 0;z<=m;z++)
        {
            dp1[0][0][z] = 0;
        }
        for(int i=1;i<=k;i++)
        {
            for(int j=0;j<=m;j++)
            {
                dp1[i][0][j] = 0;
            }
        }
        for(int i=1;i<=k;i++)
        {
            for(int j=1;j<=i;j++)
            {
                for(int z=1;z<=m;z++)
                {
                    for(int v=1;v*v<=z&&v<=j;v++)
                    {
                        dp1[i][j][z] = min(dp1[i][j][z],dp1[i-v][j-v][z-v*v]+1);
                    }
                }
            }
        }
        memset(ch,false,sizeof(ch));
        dfs(n,k,m);
        if(dp[n][k][m]>=INF)
        {
            printf("0\n");
        }else
        {
            printf("%d\n",dp[n][k][m]);
        }
    }
    return 0;
}
int dfs(int n,int m,int k)
{
    if(ch[n][m][k])
    {
        return dp[n][m][k];
    }
    if(n==1)
    {
        ch[n][m][k] = true;
        dp[n][m][k] = dp1[m][m][k];
        return dp1[m][m][k];
    }
    int Min = INF;
    for(int i=0;i<=m;i++)
    {
        if(dp1[m][i][k]!=INF)
        {
            int w = dfs(n-1,m-i,k-i);
            Min = min(Min,w+dp1[m][i][k]);
        }
    }
    ch[n][m][k] = true;
    dp[n][m][k] = Min;
    return Min;
}