UVA1442-Cav（扫描法）

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

14

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。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1000000 + 100;

int n;
int low[maxn], high[maxn];
int h[maxn];

int main()
{
    //freopen("input.txt", "r", stdin);
    int iCase;
    scanf("%d", &iCase);
    while (iCase--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &low[i]);
        }
        for (int j = 1; j <= n; j++) {
            scanf("%d", &high[j]);
        }

        int level = high[1];
        h[1] = level;
        for (int i = 2; i <= n; i++) {
            if (level > high[i]) level = high[i];
            if (level < low[i]) level = low[i];
            h[i] = level;
        }

        h[n] = min(h[n], high[n]);
        level = h[n];
        for (int i = n - 1; i >= 1; i--) {
            if (level > high[i]) level = high[i];
            if (level < low[i]) level = low[i];
            h[i] = min(h[i], level);
        }

        int ans = 0;

        for (int i = 1; i <= n; i++) {
            ans += h[i] - low[i];
        }

        printf("%d
", ans);
    }
    return 0;
}