ZOJ 3469 Food Delivery(DP)

最新推荐文章于 2020-06-06 10:17:34 发布

lab104_yifan

最新推荐文章于 2020-06-06 10:17:34 发布

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分类专栏：高效算法-动态规划

本文链接：https://blog.youkuaiyun.com/accelerator_/article/details/43069939

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本文介绍了一种使用双指针结合动态规划解决特定问题的方法。通过将输入数据分为两个集合，利用DP记录状态转移过程，实现了高效求解。文章详细展示了算法实现的代码，并解释了关键步骤。

思路：把小于x和大于x存成两个集合，然后dp[i][j][v]表示左边送完i个，右边送完j个，现在位置在左/右的代价，然后进行转移即可，具体转移见代码

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 1005;
const ll INF = 0x3f3f3f3f3f3f3f;

int n, v, x, ln, rn;
ll dp[N][N][2];

struct SB {
	int x, b;
	SB() {}
	SB(int x, int b) {
		this->x = x;
		this->b = b;
	}
} l[N], r[N];

int suml[N], sumr[N];

bool cmp1(SB a, SB b) {
	return a.x < b.x;
}

bool cmp2(SB a, SB b) {
	return a.x > b.x;
}

int main() {
	while (~scanf("%d%d%d", &n, &v, &x)) {
		ln = rn = 0;
		int a, b;
		while (n--) {
			scanf("%d%d", &a, &b);
			if (a < x) l[++ln] = SB(a, b);
			if (a > x) r[++rn] = SB(a, b);
		}
		sort(l + 1, l + ln + 1, cmp2);
		sort(r + 1, r + rn + 1, cmp1);
		l[0].x = r[0].x = x;
		for (int i = 0; i <= ln; i++)
			for (int j = 0; j <= rn; j++)
				dp[i][j][0] = dp[i][j][1] = INF;
		dp[0][0][0] = dp[0][0][1] = 0;
		for (int i = 1; i <= ln; i++)
			suml[i] = suml[i - 1] + l[i].b;
		for (int i = 1; i <= rn; i++)
			sumr[i] = sumr[i - 1] + r[i].b;
		for (int i = 0; i <= ln; i++) {
			for (int j = 0; j <= rn; j++) {
				int tot = v * (suml[ln] - suml[i] + sumr[rn] - sumr[j]);
				if (i != ln) { 
					dp[i + 1][j][0] = min(dp[i + 1][j][0], dp[i][j][0] + (l[i].x - l[i + 1].x) * tot);
					dp[i + 1][j][0] = min(dp[i + 1][j][0], dp[i][j][1] + (r[j].x - l[i + 1].x) * tot);
				}
				if (j != rn) {
					dp[i][j + 1][1] = min(dp[i][j + 1][1], dp[i][j][0] + (r[j + 1].x - l[i].x) * tot);
					dp[i][j + 1][1] = min(dp[i][j + 1][1], dp[i][j][1] + (r[j + 1].x - r[j].x) * tot);
				}
			}
		}
		printf("%lld\n", min(dp[ln][rn][0], dp[ln][rn][1]));
	}
	return 0;
}