nowcoder假日团队赛8 D-Artificial Lake
作者：互联网
链接：https://ac.nowcoder.com/acm/contest/1069/D
时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
The oppressively hot summer days have raised the cows' clamoring to its highest level. Farmer John has finally decided to build an artificial lake. For his engineering studies, he is modeling the lake as a two-dimensional landscape consisting of a contiguous sequence of N soon-to-be-submerged levels (1 <= N <= 100,000) conveniently numbered 1..N from left to right.
Each level i is described by two integers, its width W_i (1 <= W_i <= 1,000) and height (like a relative elevation) H_i (1 <= H_i <= 1,000,000). The heights of FJ's levels are unique. An infinitely tall barrier encloses the lake's model on the left and right. One example lake profile is shown below.
* * :
* * :
* * 8
* *** * 7
* *** * 6
* *** * 5
* ********** 4 <- height
* ********** 3
*************** 2
*************** 1
Level | 1 |2| 3 |
In FJ's model, he starts filling his lake at sunrise by flowing water into the bottom of the lowest elevation at a rate of 1 square unit of water per minute. The water falls directly downward until it hits something, and then it flows and spreads as room-temperature water always does. As in all good models, assume that falling and flowing happen instantly. Determine the time at which each elevation's becomes submerged by a single unit of water.
WATER WATER OVERFLOWS
| |
* | * * | * * *
* V * * V * * *
* * * .... * *~~~~~~~~~~~~*
* ** * *~~~~** : * *~~~~**~~~~~~*
* ** * *~~~~** : * *~~~~**~~~~~~*
* ** * *~~~~**~~~~~~* *~~~~**~~~~~~*
* ********* *~~~~********* *~~~~*********
*~~~~********* *~~~~********* *~~~~*********
************** ************** **************
************** ************** **************
After 4 mins After 26 mins After 50 mins
Lvl 1 submerged Lvl 3 submerged Lvl 2 submerged
Warning: The answer will not always fit in 32 bits.
输入描述:
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 describes level i with two space-separated integers: WiW_iWi and HiH_iHi
输出描述:
* Lines 1..N: Line i contains a single integer that is the number of minutes that since sunrise when level #i is covered by water of height 1.示例1
输入
3 4 2 2 7 6 4
输出
4 50 26
说明
Three levels just as in the example above. Water will fill the first level because it is the lowest.
As in the illustrations above.
模拟
每次找到当前最低的可以被灌水的平台填灌
然后合并更新
代码
1 #include <bits/stdc++.h>
2 #include <iostream>
3 #include <cstring>
4 #include <stack>
5 #include <cstdlib>
6 #include <queue>
7 #include <cmath>
8 #include <cstdio>
9 #include <algorithm>
10 #include <string>
11 #include <vector>
12 #include <list>
13 #include <iterator>
14 #include <set>
15 #include <map>
16 #include <utility>
17 #include <iomanip>
18 #include <ctime>
19 #include <sstream>
20 #include <bitset>
21 #include <deque>
22 #include <limits>
23 #include <numeric>
24 #include <functional>
25 #define gc getchar()
26 #define mem(a) memset(a,0,sizeof(a))
27 #define mod 1000000007
28 #define sort(a,n,int) sort(a,a+n,less<int>())
29 #define fread() freopen("in.in","r",stdin)
30 #define fwrite() freopen("out.out","w",stdout)
31
32 #define PI acos(-1.0)
33 #define N 100005
34 #define MOD 2520
35 #define E 1e-12
36
37 typedef long long ll;
38 typedef char ch;
39 typedef double db;
40 const long long INF = 0x3f3f3f3f3f3f3f3f;
41
42 using namespace std;
43
44
45 struct node
46 {
47 ll w , h;
48 int l , r;
49 }a[N];
50
51 int n = 0;
52 long long res[N] = {0};
53
54 int lowest()
55 {
56 ll minv = INF , pos = 0;
57 for(int i = 1;i <= n;i++)
58 if(a[i].h < minv)
59 {
60 minv = a[i].h;
61 pos = i;
62 }
63 return pos;
64 }
65 int update(int position)
66 {
67 for(int i = 0;i<n;i++)
68 {
69 int left = a[position].l , right = a[position].r;
70
71 if(a[left].h < a[position].h)
72 position = left;
73 else if(a[right].h < a[position].h)
74 position = right;
75 else
76 return position;
77 }
78 }
79 void guanshui(int position)
80 {
81 int counter = 1;
82 long long sum = 0;
83 while(counter <= N)
84 {
85 counter++;
86 sum += a[position].w;
87 res[position] = sum;
88 int l = a[position].l , r = a[position].r;
89 sum += (min(a[l].h , a[r].h) - a[position].h - 1) * a[position].w;
90
91 a[l].r = r;
92 a[r].l = l;
93
94 if(a[l].h < a[r].h)
95 {
96 a[l].w += a[position].w;
97 position = l;
98 }
99 else
100 {
101 a[r].w += a[position].w;
102 position = r;
103 }
104 position = update(position);
105 }
106 }
107 int main()
108 {
109 cin >> n;
110 int position = 0;
111 for(int i = 1;i <= n;i++)
112 {
113 cin >> a[i].w >> a[i].h;
114 }
115 for(int i = 0;i <= n+1;i++)
116 {
117 a[i].l = i-1;
118 a[i].r = i+1;
119 }
120 a[0].w = 1;
121 a[0].h = INF;
122
123 a[n+1].w = 1;
124 a[n+1].h = INF;
125
126 position = lowest();
127 guanshui(position);
128 for(int i = 1;i <= n;i++)
129 {
130 cout << res[i] << endl;
131 }
132 return 0;
133 }
标签：level,int,ll,Lake,Artificial,long,nowcoder,include,define 来源： https://www.cnblogs.com/SutsuharaYuki/p/11274832.html