HDD 6024 Building Shops (二維DP)

題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=6024

輸入不是按座標從小到大輸入的,先需要按座標從小到大排序

dp[i][j]表示前i個教室,最右邊糖果店在第j個教室時的花費

狀態轉移方程:

i != j(k\epsilon [0,i)):dp[i][j] = dp[i - 1][k] + x[i] - x[k]

i = j(k\epsilon [0,i)):dp[i][i] = dp[i - 1][k] + c[i]

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
static const int MAX_N = 3e3 + 5;
static const ll Mod = 1e9 + 7;
static const ll INF = (ll)1 << 60;
struct Node{
	int x, c;
	bool operator < (const Node &rhs) const {
		return x < rhs.x;
	}
}a[MAX_N];
ll dp[MAX_N][MAX_N];
int main() {	
	/*freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);*/
	int n;
	while (scanf("%d", &n) != EOF) {
		for (int i = 0; i < n; ++i) scanf("%d%d", &a[i].x, &a[i].c);
		sort(a, a + n);
		dp[0][0] = a[0].c;
		for (int i = 1; i < n; ++i) {
			for (int j = 0; j < i; ++j) dp[i][j] = dp[i - 1][j] + a[i].x - a[j].x;
			dp[i][i] = INF;
			for (int j = 0; j < i; ++j) dp[i][i] = min(dp[i][i], dp[i - 1][j] + a[i].c);
		}
		ll min_v = INF;
		for (int i = 0; i < n; ++i) min_v = min(min_v, dp[n - 1][i]);
		printf("%lld\n", min_v);
	}
	return 0;
}

 

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