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("%d\n", it->second - (it->first - t));
122         }
123     }
124 }

 

posted @ 2017-07-31 20:03  Angel_Kitty  阅读(380)  评论(0)  编辑  收藏  举报