POJ 1946 Cow Cycling

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探讨了在CowCycling比赛中如何通过最优策略使至少一名队员以最快速度完成比赛。介绍了比赛的基本规则，包括能量消耗、领导更替及比赛长度等关键参数。

Cow Cycling

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1479		Accepted: 788

&l t;p class="pst">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

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

/* 看似复杂的 DP 问题其实是纸老虎，只要逻辑思路对了就好写了要敢写 DP 公式啊，只要写出来了怎么实现就好商量，胆子要大啊 minT[i][j][k] 表示 i 只体力为 j 的奶牛走 k 所需的最小时间则 minT[i][j][k] = min{minT[i][j][k], minT[i-1][j-e][k-e] + minT[1][j][e]} for 1 <= e <= k and j ，表示先以速度 e 走一分钟 minT[1][j][k] = min{minT[1][j][k], minT[1][j - e * e][k - e] + 1} for 1 <= e * e <= j and e <= k 剩下只有初始化的问题了 */ #include <iostream> #define minv(a, b) ((a) <= (b) ? (a) : (b)) #define MAX_N 22 #define MAX_E 102 #define MAX_D 102 #define MAX_VAL 99999999 using namespace std; int minT[MAX_N + 1][MAX_E + 1][MAX_D + 1]; int N, E, D; int main() { int c, i, j, k; scanf("%d%d%d", &N, &E, &D); minT[1][0][0] = 0; for(i = 1; i <= E; i++) minT[1][i][0] = 0; for(i = 1; i <= D; i++) minT[1][0][i] = MAX_VAL; for(i = 1; i <= E; i++) { for(j = 1; j <= D; j++) { minT[1][i][j] = MAX_VAL; for(k = 1; k * k <= i && k <= j; k++) minT[1][i][j] = minv(minT[1][i][j], minT[1][i - k * k][j - k] + 1); } } for(c = 2; c <= N; c++) { for(i = 1; i <= E; i++) { for(j = 1; j <= D; j++) { minT[c][i][j] = MAX_VAL; for(k = 1; k <= i && k <= j; k++) minT[c][i][j] = minv(minT[c][i][j], minT[c - 1][i - k][j - k] + minT[1][i][k]); } } } if(minT[N][E][D] >= MAX_VAL) printf("0/n"); else printf("%d/n", minT[N][E][D]); return 0; }