uva 10366 - Faucet Flow

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题目大意:

A faucet is pouring water into a long, thin aquar-ium with various vertical dividers (walls) in it. Theaquarium is initially empty, and its bottom is per-fectly level. How long will it take for water to spillover its left- or right-most divider? The faucet isabove location x = 0, and the dividers are located atx= −1,−3,−5,...leftxand 1,3,5,...rightx. The di-viders are attached perpendicular to the floor and sidesof the aquarium, and have various heights. The aquar-ium’s length is greater thanrightx−leftx, its walls arehigher than the highest divider, and its width is 1 uniteverywhere. Water pours from the faucet at a rate of1 cubic unit per second.

You may assume that water is an ideal liquid: italways flows downhill and if it cannot flow downhill itspreads at an equal rate in all horizontal directions.

解题:这一题主要时注意到两边的最大值。明显又一边最大值比较大时，那么肯定是另一边先到达边界，因为这一边肯定不会到达边界。

所以这一题就是注意最大值之间的关系。

具体情况可看代码。易懂，另外附上一些测试数据

-1 3
5 5 5
-3 5
5 5 5 5 5
-7 9
1 2 3 4 1 1 1 4 5
-7 15
1 2 3 4 1 1 1 4 4 4 4 4
-9 7
5 4 1 1 1 4 3 2 1
-15 7
4 4 4 4 4 1 1 1 4 3 2 1
-7 7
4 3 2 1 1 2 3 4
-7 7
4 3 2 1 1 2 2 4
-7 7
4 3 1 1 1 2 3 4
-7 9
4 3 2 1 1 2 3 4 1
-9 7
1 4 3 2 1 1 2 3 4
-7 9
2 1 1 5 5 1 1 1 2
-1 1
1 1
-1 1
2 1
-7 5
1 1 1 2 3 4 3
-11 9
1 3 5 7 9 10 10 8 6 4 2
-9 9
1 4 3 2 1 1 2 3 4 1
-9 11
1 4 3 2 1 1 2 3 4 1 2
-11 9
2 1 4 3 2 1 1 2 3 4 1
-3 15
19 20 20 1 1 1 1 1 1 1
-3 15
19 20 20 3 3 3 3 3 3 3
-15 3
1 1 1 1 1 1 1 20 20 19
-15 3
3 3 3 3 3 3 3 20 20 19
-15 3
8 3 8 3 3 7 2 1 7 2
-9 9
5 5 8 7 4 2 8 6 10 7
-9 9
7 10 6 8 2 4 7 8 5 5
-3 7
10 4 4 4 2 1
-11 7
10 2 2 2 2 4 4 4 2 1

答案:

100

116

104

////  main.cpp//  uva 10366 - Faucet Flow////  Created by XD on 15/8/19.//  Copyright (c) 2015年 XD. All rights reserved.//#include <iostream>#include <string>#include <queue>#include <stack>#include <stdio.h>#include <stdlib.h>#include <math.h>#include<vector>#include <string.h>#include <string>#include <algorithm>#include <set>#include <map>#include <cstdio>#define ll long longusing namespace std  ;const int maxn  =1000 + 10 ;struct divider{    int high  ;    int pos  ;};int wall_left[maxn] ;int wall_right[maxn] ;int  l ,r ,n;ll min(ll x , ll y){    return x < y ? x :y ;}int main(int argc, const char * argv[]) {    ll ans = 0  ;    while (scanf("%d%d" ,&l,&r)==2&&r!=0&&l!=0) {        ans = 0 ;        //输入左边的墙的高度        int len1=( -l + 1)/2 ;        int max1 = 0 ;int pos1= 0  ;        ll ans1 = 0 ;ll temp_max  = 0 ;        for (int i = 0; i < len1; i++) {            scanf("%d" ,&wall_left[i]) ;            if (max1 <=wall_left[i]) {                max1 = wall_left[i] ;                pos1 = i ;            }        }        //统计越左边的最高的墙的左边边界需要的水的体积的二倍        temp_max = wall_left[0] ;        for (int i = 1; i <=pos1; i++) {            if (wall_left[i] <= temp_max) {                ans1+=(temp_max * 4) ;            }            else{                ans1+=(temp_max*4) ;                temp_max = wall_left[i] ;            }        }         //输入左边的墙的高度        int len2 = (r +1)/2  ;        int max2 = 0 ;int pos2 = 0 ;ll ans2 = 0 ;        for (int i = 0  ;  i < len2; i++) {            scanf("%d" ,&wall_right[i]) ;            if (max2 <wall_right[i]) {                max2 = wall_right[i] ;                pos2 = i ;            }        }         //统计右边的最高的墙的右边边界需要的水的体积的二倍        temp_max = wall_right[len2-1] ;        for (int i = len2-2; i >=pos2;   i--) {            if (wall_right[i] <= temp_max) {                ans2+=(temp_max * 4) ;            }            else{                ans2+=(temp_max*4) ;                temp_max = wall_right[i] ;            }        }        //左右两边的最高墙之间的关系。        /*         这里主要注意一种情况就是  当有一边的最大值比较大（假设为右边）时，要注意右边的第一个等于max1的和第一个大于max1的。         */        if (max1> max2) {                        int tpos1 = -1 ;int tpos2 = -1 ;            for (int i = len1-1; i>=pos1 ; i--) {                if (wall_left[i]==max2) {                    if (tpos1==-1) {                       tpos1 = i ;                    }                }                if (wall_left[i] > max2) {                    tpos2 = i ;                    break ;                }            }            if (tpos1 == -1) {                ans = max2  *(len1 - tpos2 + pos2) *2 + ans2/2 ;            }            else{                ll ans3 = max2 * (tpos1 - tpos2) *2   ;                if (ans3 > ans2/ 2) {                    ans = max2 * (len1 - tpos1 + pos2)*2 + ans2 ;                }                else{                    ans = max2 * (len1 - tpos1 + pos2)*2 + ans3 * 2 +(ans2/2 - ans3) ;                }            }        }        else if ( max1 == max2)        {            ans =max2 * (pos2 + len1 - pos1 )  *2 ;            ans =ans +  min(ans1 , ans2) ;        }        else{            int tpos1 = -1 ;int tpos2 = -1 ;            for (int i = 0; i<=pos2 ; i++) {                if (wall_right[i]==max1) {                   if(tpos1==-1 )tpos1 = i ;                    continue ;                }                if (wall_right[i] > max1) {                    tpos2 = i ;                    break ;                }            }            if (tpos1 == -1) {                ans = max1  *(tpos2 + len1 -  pos1 ) *2 + ans1/2 ;            }            else{                ll ans3 = max1 * (tpos2 - tpos1) *2   ;                if (ans3 > ans1/ 2) {                    ans = max1 * (len1 - pos1 + tpos1)*2 + ans1 ;                }                else{                    ans = max1 * (len1 - pos1 + tpos1)*2 + ans3 * 2 +(ans1/2 - ans3) ;                }            }        }        printf("%lld\n",ans) ;    }    return 0;}

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