Codeforces 833E Caramel Clouds
E. Caramel Clouds
time limit per test:3 seconds
memory limit per test:256 megabytes
input:standard input
output:standard output
It is well-known that the best decoration for a flower bed in Sweetland are vanilla muffins. Seedlings of this plant need sun to grow up. Slastyona has m seedlings, and the j-th seedling needs at least kj minutes of sunlight to grow up.
Most of the time it's sunny in Sweetland, but sometimes some caramel clouds come, the i-th of which will appear at time moment (minute) li and disappear at time moment ri. Of course, the clouds make shadows, and the seedlings can't grow when there is at least one cloud veiling the sun.
Slastyona wants to grow up her muffins as fast as possible. She has exactly C candies, which is the main currency in Sweetland.
One can dispel any cloud by paying ci candies. However, in order to comply with Sweetland's Department of Meteorology regulations, one can't dispel more than two clouds.
Slastyona hasn't decided yet which of the m seedlings will be planted at the princess' garden, so she needs your help. For each seedling determine the earliest moment it can grow up if Slastyona won't break the law and won't spend more candies than she has. Note that each of the seedlings is considered independently.
The seedlings start to grow at time moment 0.
Input
The first line contains two integers n and C (0 ≤ n ≤ 3·105, 0 ≤ C ≤ 109) – the number of caramel clouds and the number of candies Slastyona has.
The next n lines contain three integers each: li, ri, ci (0 ≤ li < ri ≤ 109, 0 ≤ ci ≤ 109), describing one caramel cloud.
The next line contains single integer m (1 ≤ m ≤ 3·105) – the number of seedlings. Each of the seedlings is described with one integer kj (1 ≤ kj ≤ 109) – the required number of sunny minutes.
Output
For each seedling print one integer – the minimum minute Slastyona can grow it up.
Examples
Input
3 5
1 7 1
1 6 2
1 7 1
3
7
2
5
Output
12
7
10
Input
3 15
1 4 17
2 8 6
4 8 9
2
5
1
Output
8
1
Input
2 10
3 7 9
10 90 10
2
10
100
Output
10
104
Note
Consider the first example. For each k it is optimal to dispel clouds 1 and 3. Then the remaining cloud will give shadow on time segment [1..6]. So, intervals [0..1] and [6..inf) are sunny.
In the second example for k = 1 it is not necessary to dispel anything, and for k = 5 the best strategy is to dispel clouds 2 and 3. This adds an additional sunny segment [4..8], which together with [0..1] allows to grow up the muffin at the eight minute.
If the third example the two seedlings are completely different. For the first one it is necessary to dispel cloud 1 and obtain a sunny segment [0..10]. However, the same strategy gives answer 180 for the second seedling. Instead, we can dispel cloud 2, to make segments [0..3] and [7..inf) sunny, and this allows up to shorten the time to 104.
题目链接:http://codeforces.com/problemset/problem/833/E
叉姐的题解:
叉姐的代码:
1 #include <algorithm>
2 #include <cstdio>
3 #include <cstring>
4 #include <map>
5 #include <set>
6 #include <utility>
7 #include <vector>
8
9 const int N = 300000;
10
11 struct Sum
12 {
13 int add(int id, int value)
14 {
15 if (a[0].second == id) {
16 a[0].first = std::max(a[0].first, value);
17 } else if (a[1].first < value) {
18 a[1] = {value, id};
19 }
20 if (a[0].first < a[1].first) {
21 std::swap(a[0], a[1]);
22 }
23 }
24
25 int ask(int id)
26 {
27 if (a[0].second != id) {
28 return a[0].first;
29 }
30 return a[1].first;
31 }
32
33 std::pair<int, int> a[2] = {{0, -1}, {0, -1}};
34 };
35
36 int cost[N + 1], toupd[N];
37
38 int main()
39 {
40 #ifdef LOCAL_JUDGE
41 freopen("E.in", "r", stdin);
42 #endif
43 int n, budget;
44 while (scanf("%d%d", &n, &budget) == 2) {
45 cost[n] = 0;
46 std::vector<std::pair<int, int>> events;
47 events.emplace_back(0, n);
48 events.emplace_back(2000000000, n);
49 for (int i = 0, l, r; i < n; ++ i) {
50 scanf("%d%d%d", &l, &r, cost + i);
51 events.emplace_back(l, i);
52 events.emplace_back(r, i);
53 }
54 std::sort(events.begin(), events.end());
55 std::vector<int> values(cost, cost + n);
56 std::sort(values.begin(), values.end());
57 values.erase(std::unique(values.begin(), values.end()), values.end());
58 std::set<int> covers;
59 if (events[0].second < n) {
60 covers.insert(events[0].second);
61 }
62 int curmx = 0;
63 std::vector<std::pair<int, int>> parts;
64 memset(toupd, 0, sizeof(toupd));
65 std::vector<Sum> bit(values.size());
66 std::map<std::pair<int, int>, int> length;
67 for (int t = 1; t < (int)events.size(); ++ t) {
68 int mxlen = events[t].first - events[t - 1].first;
69 if (mxlen > 0 && (int)covers.size() <= 2) {
70 int p = n, q = n;
71 if ((int)covers.size() > 0) {
72 p = *covers.begin();
73 }
74 if ((int)covers.size() > 1) {
75 q = *covers.rbegin();
76 }
77 int start = -1;
78 if (p == n) { // 0
79 start = curmx;
80 } else {
81 if (q == n) { // 1
82 if (cost[p] <= budget) {
83 start = toupd[p];
84 for (int k = (int)(std::upper_bound(values.begin(), values.end(), budget - cost[p]) - values.begin()) - 1; k >= 0; k -= ~k & k + 1) {
85 start = std::max(start, bit[k].ask(p));
86 }
87 auto value = length[{p, q}] + mxlen;
88 for (int k = std::lower_bound(values.begin(), values.end(), cost[p]) - values.begin(); k < (int)values.size(); k += ~k & k + 1) {
89 bit[k].add(p, value);
90 }
91 }
92 } else if (cost[p] + cost[q] <= budget) {
93 start = length[{p, n}] + length[{q, n}];
94 toupd[p] = std::max(toupd[p], length[{q, n}] + length[{p, q}] + mxlen);
95 toupd[q] = std::max(toupd[q], length[{p, n}] + length[{p, q}] + mxlen);
96 }
97 if (~start) {
98 start += length[{p, q}] + length[{n, n}];
99 }
100 }
101 if (~start && start + mxlen > curmx) {
102 curmx = start + mxlen;
103 parts.emplace_back(curmx, events[t].first);
104 }
105 length[{p, q}] += mxlen;
106 }
107 auto&& i = events[t].second;
108 if (i < n) {
109 if (covers.count(i)) {
110 covers.erase(i);
111 } else {
112 covers.insert(i);
113 }
114 }
115 }
116 int q, t;
117 scanf("%d", &q);
118 while (q --) {
119 scanf("%d", &t);
120 auto it = std::lower_bound(parts.begin(), parts.end(), std::make_pair(t, 0));
121 printf("%dn", it->second - (it->first - t));
122 }
123 }
124 }
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法
- 一起来学演化计算-实数空间变异算子
- 卡特兰数入门
- 常见编程模式之动态规划:0-1背包问题
- stat 命令家族(2)- 详解 pidstat
- MTO和MaTO MMZDT
- stat 命令家族(3)- 详解 mpstat
- 知识图谱入门(一)
- PHP判断变量内容是什么编码(gbk?utf-8) mb_detect_encoding
- stat 命令家族(4)- 详解 iostat
- PHP将数组存入数据库中的四种方式
- 序列化与json性能评测
- js内存泄漏常见的四种情况(From LeuisKen)
- 「R」Rprofile:R 全局设置
- Jmetal Problem和Problem Set的变量范围
- 简单工厂、工厂方法、抽象工厂的比较与分析