1442 - Cav

As an owner of a land with a cave you weredelighted when you last heard that underground fuel tanks are great business.Of course, the more volume one can store, the better. In case of your cave, theefflective volume is not easy to calculate, because the cave has a rathersophisticated shape (see figure). Thank heavens it is degenerate in onedimension!

The cave. All ponds that can be floodedwith fuel are marked black.

Furthermore, there is some electricalwiring on the ceiling of the cave. You can never be sure if the insulation isintact, so you want to keep the fuel level just below the ceiling at everypoint. You can pump the fuel to whatever spots in the cave you choose, possiblycreating several ponds. Bear in mind though that the fuel is a liquid, so itminimises its gravitational energy, e.g., it will run evenly in every directionon a flat horizontal surface, pour down whenever possible, obey the rule ofcommunicating vessels, etc. As the cave is degenerate and you can make thespace between the fuel level and the ceiling arbitrarily small, you actuallywant to calculate the maximum possible area of ponds that satisfyaforementioned rules.

Input 

The input contains several test cases. Thefirst line of the input contains a positive integerZ15, denoting the number oftest cases. Then Z test cases follow, each conforming to theformat described below

In the first line of an input instance, there is an integer n (1n10⁶) denotingthe width of the cave. The second line of input consists of n integers p₁, p₂,..., p_n andthe third line consists of n integers s₁, s₂,..., s_n,separated by single spaces. The numbers p_i and s_isatisfy 0p_i < s_i1000 and denote thefloor and ceiling level at interval [i, i + 1),respectively.

Output 

For each test case, your program has towrite an output conforming to the format described below.

Your program is to print out one integer: the maximum total area of admissibleponds in the cave.

SampleInput 

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

SampleOutput 

14

代码:

#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1000000 + 5;

int n, p[maxn], s[maxn], h[maxn];

int main() {

int T;

scanf("%d", &T);

while(T--) {

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 ans = 0, level = s[0];

for(int i = 0; i < n; i++) {

if(p[i] > level) level = p[i];

if(s[i] < level) level = s[i];

h[i] = level;

}

level = s[n-1];

for(int i = n-1; i >= 0; i--) {

if(p[i] > level) level = p[i];

if(s[i] < level) level = s[i];

ans += min(h[i], level) - p[i];

}

printf("%d\n", ans);

}

return 0;

}