HDU - 4526 威威猫系列故事――拼车记 （DP）

 14 
思路：设dp[i][j]表示前i辆车还剩j个人的最小花费。那么我们需要去每次枚举减少的人
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 110;
const int inf = 0x3f3f3f3f;

int dp[maxn][maxn];
int n, k, s, d;

int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d%d%d", &n, &k, &d, &s);
		for (int i = 0; i <= k; i++)
			for (int j = 0; j <= n; j++)
				dp[i][j] = inf;

		dp[0][n] = 0;
		int last = 0;
		for (int i = 1; i <= k; i++) {
			int t, p;
			scanf("%d%d", &t, &p);
			for (int j = n; j >= 0; j--) {
				if (dp[i-1][j] != inf) {
					dp[i][j] = min(dp[i][j], dp[i-1][j] + j * (t - last));
					for (int l = 1; l <= p; l++)
						dp[i][j-l] = min(dp[i][j-l], dp[i-1][j] + j * (t - last) + d);
				}
			}
			last = t;
		}

		if (dp[k][0] == inf)
			printf("impossible\n");
		else printf("%d\n", dp[k][0]);
	}
	return 0;
}