POJ 1661 Help Jimmy(递推DP)

思路:

1. 每个板子有左右两端, dp[i][0], dp[i][1] 分别记录左右端到地面的时间

2. 从下到上递推计算, 上一层的板子必然会落到下面的某一层板子上, 或者地面上

总结:

1. 计算每个板子的 dp[i][0/1] 仅需考虑该板子的直接前驱即可

2. 动规的思想并不很明显

3. 代码中, 两个板子相对位置的判断特别精髓

4. 将地面和初始状态都抽象成一块板子

代码:

#include <iostream>
#include <algorithm>
using namespace std;

class board {
public:
	int x1, x2, h;
	board(int _x1, int _x2, int _h):x1(_x1), x2(_x2), h(_h){}
	board() {
		board(-1,-1,-1);
	}
	bool operator <(const board & other) const {
		return this->h < other.h;
	}
};

const int INF = 0X3F3F3F3F;
const int MAXN = 1010;
int t, N, X, Y, H, MAX;
board boards[MAXN];
int dp[MAXN][2];

int mainFunc() {

	for(int i = 0; i <= N+1; i ++) {
		for(int j = i-1; j >= 0; j --) {
			if(boards[i].x1 >= boards[j].x1 && boards[i].x1 <= boards[j].x2) { // i 的左端可以掉落到 j 上
				int h = boards[i].h - boards[j].h;
				if(h > MAX) dp[i][0] = INF;
				else if (j == 0) dp[i][0] = h;
				else
					dp[i][0] = min(dp[j][0]+boards[i].x1-boards[j].x1, dp[j][1]+boards[j].x2-boards[i].x1) + h;
				break;
			}
		}
		for(int j = i-1; j >= 0; j --) {
			if(boards[i].x2 >= boards[j].x1 && boards[i].x2 <= boards[j].x2) { // i　的右端可以掉到　j 上
				int h = boards[i].h - boards[j].h;
				if(h > MAX) dp[i][1] = INF;
				else if(j == 0) dp[i][1] = h;
				else
					dp[i][1] = min(dp[j][0]+boards[i].x2-boards[j].x1, dp[j][1]+boards[j].x2-boards[i].x2) + h;
				break;
			}
		}
	}
	return dp[N+1][1];
}

int main() {
	freopen("E:\\Copy\\ACM\\poj\\1661\\in.txt", "r", stdin);
	cin >> t;
	while(t-- >= 1) {
		cin >> N >> X >> H >> MAX;
		for(int i = 0; i < N; i ++) {
			cin >> boards[i].x1 >> boards[i].x2 >> boards[i].h;
		}
		boards[N].x1 = -20010, boards[N].x2 = 20010, boards[N].h = 0;
		boards[N+1].x1 = X, boards[N+1].x2 = X, boards[N+1].h = H;
		sort(boards, boards+N+2);
		// mainFunction
		cout << mainFunc() << endl;
	}
	return 0;
}

　　

update 2014年3月16日10:36:58

1. 直接前驱可以预处理得到