UVA1442 Cave

题解：

从左往右扫描，初始时设水位level=s[0]，然后依次判断各个位置[i,i+1]处的高度

1. 如果p[i] > level，说明水被隔绝了，需要把level提升到pi

2. 如果s[i] < level，说明水位太高，碰到天花板，需要把level降低到si

3. 位置[i,i+1]处的水位就是扫描到位置i时的level

代码

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 100000
#define LL long long
int cas=1,T;
int p[1000005];
int s[1000005];
int h[1000005];

int main()
{
scanf("%d",&T);
while (T--)
{
int n;
scanf("%d",&n);
for (int i = 0;i<n;i++)
scanf("%d",&p[i]);
for (int i = 0;i<n;i++)
scanf("%d",&s[i]);
int level = s[0];
for (int i = 0;i<n;i++)
{
if (level>s[i])
level = s[i];
if (level < p[i])
level = p[i];
h[i]=level;
}

level = s[n-1];
int ans =0;
for (int i = n-1;i>=0;i--)
{
if (level >s[i])
level = s[i];
if (level < p[i])
level = p[i];
ans+=min(level,h[i])-p[i];
}
printf("%d\n",ans);
}
return 0;
}