UVA1442-Cav（扫描法）

最新推荐文章于 2024-03-27 23:42:39 发布

转载最新推荐文章于 2024-03-27 23:42:39 发布 · 272 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/npugen/p/9678492.html

本文介绍了一种解决UVA1442-Cav问题的有效算法——扫描法。通过两次扫描（从左到右和从右到左），确定了洞穴中能够形成的最大面积的水塘。具体步骤包括初始化最高水平面、考虑墙壁限制条件等。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem UVA1442-Cav

Accept: 185 Submit: 679
Time Limit: 3000 mSec

Problem Description

Input

The input contains several test cases. The ﬁrst line of the input contains a positive integer Z ≤ 15, denoting the number of test cases. Then Z test cases follow, each conforming to the format described below. In the ﬁrst line of an input instance, there is an integer n (1 ≤ n ≤ 106) denoting the width of the cave. The second line of input consists of n integers p1,p2,...,pn and the third line consists of n integers s1,s2,...,sn, separated by single spaces. The numbers pi and si satisfy 0 ≤ pi < si ≤ 1000 and denote the ﬂoor and ceiling level at interval [i,i + 1), respectively.

Output

For each test case, your program is to print out one integer: the maximum total area of admissible ponds in the cave.

Sample Input

1
15
6 6 7 5 5 5 5 5 5 1 1 3 3 2 2
10 10 10 11 6 8 7 10 10 7 6 4 7 11 11

Sample Output

题解：扫描法。对于每一个单位小区间[i,i+1],它的最高高度应该满足，在这个高度向左右两边发出两条射线，这两条射线应穿出洞穴或者碰到下面的墙壁，这样一来，我们就可以用扫描法从左到右和从右到左分别进行扫描，从左到右扫描时只看左边的射线，右边同理。扫描的过程，其实就是在维护一个最高高度，由物理的连通器原理，当前单位小区间的高度不应高于左边的小区间，但是如果该区间和左边不连通，那就可以大于左边的高度，这种情况也就是该处墙壁最低点高于左边水位，如果当前高度大于上面的墙壁，那么肯定要降低到上面墙壁的高度，利用这样两条规则就可以扫描出合理的最高高度。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 1000000 + 100;
 6 
 7 int n;
 8 int low[maxn], high[maxn];
 9 int h[maxn];
10 
11 int main()
12 {
13     //freopen("input.txt", "r", stdin);
14     int iCase;
15     scanf("%d", &iCase);
16     while (iCase--) {
17         scanf("%d", &n);
18         for (int i = 1; i <= n; i++) {
19             scanf("%d", &low[i]);
20         }
21         for (int j = 1; j <= n; j++) {
22             scanf("%d", &high[j]);
23         }
24 
25         int level = high[1];
26         h[1] = level;
27         for (int i = 2; i <= n; i++) {
28             if (level > high[i]) level = high[i];
29             if (level < low[i]) level = low[i];
30             h[i] = level;
31         }
32 
33         h[n] = min(h[n], high[n]);
34         level = h[n];
35         for (int i = n - 1; i >= 1; i--) {
36             if (level > high[i]) level = high[i];
37             if (level < low[i]) level = low[i];
38             h[i] = min(h[i], level);
39         }
40 
41         int ans = 0;
42 
43         for (int i = 1; i <= n; i++) {
44             ans += h[i] - low[i];
45         }
46 
47         printf("%d\n", ans);
48     }
49     return 0;
50 }