UVa1025

UVa1025（dp）

紫书的dp入门题目，属于DAG上求最短路径

状态d(i,j)表示i时刻人在车站j的话还需等待的最少时间

对于每种状态有2种决策：

1：无车：等待1分钟

2：有车：搭乘地铁前往最近的一站（左或右选一方向）

数组d的i维需要多开一些，因为d[i + time[j-1]]

可能造成数组溢出引发RE

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

int main() {
	int n;
	int m1, m2;
	int kase = 0;
	int t;

	while (scanf("%d", &n) != EOF&&n) {
		kase++;
		int has_train[302][52][2] = { 0 };
		int d[302][52] = { 0 };
		int time[52] = { 0 };	
		
		scanf("%d", &t);

		for (int i = 1; i < n; i++) {
			scanf("%d", &time[i]);
		}

		scanf("%d", &m1);
		for (int i = 0; i < m1; i++) {
			int tem;
			scanf("%d", &tem);
			int a = tem;
			for (int j = 1; j <= n; j++) {
				has_train[a][j][0] = 1;
				a += time[j];
			}
		}

		scanf("%d", &m2);
		for (int i = 0; i < m2; i++) {
			int tem;
			scanf("%d", &tem);
			int a = tem;
			for (int j = n; j >= 1; j--) {
				has_train[a][j][1] = 1;
				a += time[j - 1];
			}
		}


		for (int i = 1; i <= n - 1; i++) { d[t][i] = inf; }
		d[t][n] = 0;

		for (int i = t - 1; i >= 0; i--) {					//d[i][j]表示i时刻，在j站需等的最少时间
			for (int j = 1; j <= n; j++) {
				d[i][j] = d[i + 1][j] + 1;
				if (j < n&&has_train[i][j][0] && i + time[j] <= t) {
					d[i][j] = min(d[i][j], d[i + time[j]][j + 1]);
				}
				if(j > 1&&has_train[i][j][1]&&i+time[j - 1]<=t){
					d[i][j] = min(d[i][j], d[i + time[j - 1]][j - 1]);
				}

			}
		
		}

		cout << "Case Number " << kase << ": ";
		if (d[0][1] >= inf) cout << "impossible\n";
		else cout << d[0][1] << "\n";
	}


}