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  • 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
    ";
    		else cout << d[0][1] << "
    ";
    	}
    
    
    }


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  • 原文地址:https://www.cnblogs.com/Drenight/p/8611339.html
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