PKU 1946 - Cow Cycling (DP)
题意
几只牛在赛跑，每一只牛带头跑x圈/分需要消耗x*x的体力，现在问最少需要多少分。
思路
显然是让最后一只冲刺。
dp[i][j][k]
表示第i只牛跑j圈用k能量的最少分钟数。然后枚举他上一口气跑的圈数。
第
i+1
只牛继承第i只牛的成果。即
dp[i+1][j][j] = dp[i][j][k]
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int MAXN = 1e6 + 10;
const int MOD = 9901;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
int dp[25][110][110];
int main()
{
int N, E, D;
while (~scanf("%d%d%d", &N, &E, &D))
{
MS(dp, INF);
dp[1][0][0] = 0;
for (int i = 1; i <= N; i++)
for (int j = 1; j <= D; j++)
for (int k = 1; k <= E; k++)
{
for (int l = 1; l*l <= k && l <= j; l++)
dp[i][j][k] = min(dp[i][j][k], dp[i][j-l][k-l*l]+1);
dp[i+1][j][j] = min(dp[i+1][j][j], dp[i][j][k]);
}
int ans = INF;
for (int i = 1; i <= E; i++) ans = min(ans, dp[N][D][i]);
printf("%d\n", ans);
}
return 0;
}