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本文链接：https://blog.youkuaiyun.com/fp_hzq/article/details/6244831

Faucet Flow

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 276		Accepted: 81

Description

A faucet is pouring water into a long, thin aquarium with various vertical dividers (walls) in it. The aquarium is initially empty, and its bottom is perfectly level. How long will it take for water to spill over its left- or right-most divider? The faucet is above location x=0, and the dividers are located at x=-1, -3, -5, ..., leftx and 1, 3, 5, ..., rightx. The dividers are attached perpendicular to the floor and sides of the aquarium, and have various heights. The aquarium's length is greater than rightx-leftx, its walls are higher than the highest divider, and its width is 1 unit everywhere. Water pours from the faucet at a rate of 1 cubic unit per second. [You may assume that water is an ideal liquid: it always flows downhill and if it cannot flow downhill it spreads at an equal rate in all horizontal directions.]

Input

Each test case consists of two integers leftx (an odd number <= -1) and rightx (an odd number >= 1). Subsequent lines contain the height (a positive integer) of each divider from left to right. There will be no more than 1000 dividers in any test case. Input is terminated with a line containing two zeros.

Output

For each case, output an integer on a single line, indicating how long it will take, in seconds, before water starts spilling over either the left or right divider.

Sample Input

-1 13 5-3 34 3 2 1-3 51 2 2 1 10 0

Sample Output

668

Source

Waterloo local 2002.09.21

分析：逻辑问题。。。做两个阶梯，一个从左上升，一个从右上升，判断水龙头下方的那个区间哪个阶梯高，

1。一样高：选择小的那边乘二，再加上中间共有的那部分

2。右边高：找出最靠左边的比左边阶梯高的那个挡板，第一个一样高的要特判并求那段水注满的时间，

之后类似情况一

3。同情况二

#include <iostream>
using namespace std;
int main()
{
    //freopen("car.in","r",stdin);
    //freopen("car.out","w",stdout);
    int m,n,a[1010],f[1010],g[1010];
    while(scanf("%d%d",&m,&n),(m!=0)&&(n!=0))
    {
        int s=0,l,r,now;
        for(int i=m;i<=n;i+=2)
        {
            ++s;
            scanf("%d",&a[s]);
            if(i==-1)l=s;
            if(i==1)r=s;
            f[s]=g[s]=0;
        }
        now=g[1]=1;
        for(int i=2;i<r;++i)
            if(a[i]>=a[now])f[i]=f[i-1]+(i-now)*2*a[now],g[i]=now=i;
            else f[i]=f[i-1],g[i]=now;
        now=f[s]=s;
        for(int i=s-1;i>l;--i)
            if(a[i]>=a[now])g[i]=g[i+1]+(now-i)*2*a[now],f[i]=now=i;
            else g[i]=g[i+1],f[i]=now;
        if(a[g[l]]==a[f[r]])printf("%d/n",min(f[l],g[r])*2+(f[r]-g[l])*2*a[g[l]]);
        else if(a[g[l]]<a[f[r]])
        {
            now=0;
            bool ok=false;
            while(a[g[l]]>=a[r])
            {
                if((!ok)&&(a[g[l]]==a[r]))now+=(r-g[l])*2*a[r],g[l]=r,ok=true;
                ++r;
            }
            if((ok)&&(f[l]<=(r-g[l])*2*a[g[l]]))printf("%d/n",f[l]*2+now);
            else printf("%d/n",f[l]+now+(r-g[l])*2*a[g[l]]);
        }
        else
        {
            now=0;
            bool ok=false;
            while(a[f[r]]>=a[l])
            {
                if((!ok)&&(a[f[r]]==a[l]))now+=(f[r]-l)*2*a[l],f[r]=l,ok=true;
                --l;
            }
            if((ok)&&(g[r]<=(f[r]-l)*2*a[f[r]]))printf("%d/n",g[r]*2+now);
            else printf("%d/n",(f[r]-l)*2*a[f[r]]+now+g[r]);
        }
    }
    return 0;
}