【高效算法设计——双向扫描】 UVa 1442 Cave

Cav

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description





As an owner of a land with a cave you were delighted 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, the efflective
volume is not easy to calculate, because the cave has a rather sophisticated shape (see figure). Thank heavens it is degenerate in one dimension!



The cave. All ponds that can be flooded with fuel are marked black.
Furthermore, there is some electrical wiring on the ceiling of the cave. You can never be sure if the insulation is intact, so you want to keep the fuel level just below the ceiling at every point. You can pump
the fuel to whatever spots in the cave you choose, possibly creating several ponds. Bear in mind though that the fuel is a liquid, so it minimises its gravitational energy, e.g., it will run evenly in every direction on a flat horizontal surface, pour down
whenever possible, obey the rule of communicating vessels, etc. As the cave is degenerate and you can make the space between the fuel level and the ceiling arbitrarily small, you actually want to calculate the maximum possible area of ponds that satisfy aforementioned
rules.

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 Ztest 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 nintegers 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 has to write an output conforming to the format described below.

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

思路：假设x坐标为i的点，底部高度为s[i],顶部为p[i]，那么该点的燃料高度，应该为以它为中心向左右延伸直到碰壁所能达到的最大高度，假设L[i]表示往左能到达的最大高度，R[i]表示往右能达到的最大高度，那么第i点的燃料储备量即min(L[i],R[i])-s[i]

L[i]和R[i]的求法相同，只是扫描的方向相反而已，比如求L[i]，我们先把L[0]设为p[0],当前高度为L[i-1]为h，如果h大于p[i]，则h为p[i]，如果h小与s[i]，则h为s[i]，代码如下

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<map>
using namespace std;

const int maxn=1000000+5;

int s[maxn],p[maxn],level[maxn];
int n;

inline bool get(int &t)
{
    bool flag = 0 ;
    char c;
    while(!isdigit(c = getchar())&&c!='-') if( c == -1 ) break ;
    if( c == -1 ) return 0 ;
    if(c=='-') flag = 1 , t = 0 ;
    else t = c ^ 48;
    while(isdigit(c = getchar()))    t = (t << 1) + (t << 3) + (c ^ 48) ;
    if(flag) t = -t ;
    return 1 ;
}

int main()
{
    int T,i;
    get(T);
    while(T--)
    {
        get(n);
        for(i=0;i<n;i++)
            get(s[i]);
        for(i=0;i<n;i++)
            get(p[i]);
        int h;
        h=p[0];
        for(i=0;i<n;i++)
        {
            if(p[i]<h)
                h=p[i];
            if(s[i]>h)
                h=s[i];
            level[i]=h;

        }

        h=p[n-1];
        int ans=0;
        for(i=n-1;i>=0;i--)
        {
            if(p[i]<h)
                h=p[i];
            if(s[i]>h)
                h=s[i];

            ans+=min(h,level[i])-s[i];

        }

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