POJ 1946 Cow Cycling
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 1479 | Accepted: 788 |
&l t;p class="pst">Description
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
Output
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
/* 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; }