毁灭者问题

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题目及解题代码存于此处以便研读

样例输入

5
0 10 1
0 12 1
0 20 1
0 12 1
0 10 1
2
5 1 5
19 1 5

样例输出

83


AC的代码

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #include <string.h>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <map>
  8 #include <set>
  9 #include <queue>
 10 #include <bitset>
 11 #include <string>
 12 #define PQ priority_queue
 13 #define OO 2147483647
 14 #define Max(a, b) ((FASTBUFFER = ((a) - (b)) >> 31), ((b) & FASTBUFFER | (a) & ~FASTBUFFER))
 15 #define Min(a, b) ((FASTBUFFER = ((a) - (b)) >> 31), ((a) & FASTBUFFER | (b) & ~FASTBUFFER))
 16 #define Swap(a, b) (a ^= b, b ^= a, a ^= b)
 17 
 18 using namespace std;
 19 
 20 const int N = 100005;
 21 
 22 typedef long long ll;
 23 
 24 inline int ran() {
 25     static int x = 1;
 26     x += (x << 1) + 1;
 27     return x & 2147483647;
 28 }
 29 
 30 struct Node;
 31 
 32 typedef pair <Node*, Node*> Pair;
 33 
 34 Node *null;
 35 
 36 struct Node {
 37     int val, snow, size;
 38     ll sum;
 39     Node *left, *right;
 40 
 41     Node (int val, int snow, Node *left, Node *right) : 
 42     val(val), snow(snow), size(snow), left(left), right(right), sum((ll)val * snow) {}
 43 
 44     Node *Update() {
 45         size = left->size + snow + right->size;
 46         sum = left->sum + (ll)val * snow + right->sum;
 47         return this;
 48     }
 49 
 50     Pair split(int v);
 51 };
 52 
 53 Node *Merge(Node *a, Node *b) {
 54     if (a == null) {
 55         return b;
 56     }
 57 
 58     if (b == null) {
 59         return a;
 60     }
 61 
 62     if (ran() % (a->size + b->size) < a->size) {
 63         a->right = Merge(a->right, b);
 64         return a->Update();
 65     }
 66 
 67     b->left = Merge(a, b->left);
 68     return b->Update();
 69 }
 70 
 71 Pair Node :: split(int v) {
 72     if (this == null) {
 73         return make_pair(null, null);
 74     }
 75 
 76     if (val >= v) {
 77         Pair ret = left->split(v);
 78         left = ret.second;
 79         return make_pair(ret.first, this->Update());
 80     }
 81 
 82     Pair ret = right->split(v);
 83     right = ret.first;
 84     return make_pair(this->Update(), ret.second);
 85 }
 86 
 87 Node *root;
 88 
 89 struct monsterNode {
 90     int s, m, r;
 91 }a[N];
 92 
 93 int n, m;
 94 ll ans;
 95 multiset <int> s;
 96 vector <int> listInsert[N], listErase[N];
 97 
 98 void insertWithTreap(int v) {
 99     Pair ret1 = root->split(v), ret2 = ret1.second->split(v + 1);
100     if (ret2.first->size) {
101         ret2.first->snow++;
102         ret2.first->size++;
103         ret2.first->sum += ret2.first->val;
104         root = Merge(ret1.first, Merge(ret2.first, ret2.second));
105         return;
106     }
107 
108     root = Merge(ret1.first, Merge(new Node(v, 1, null, null), ret2.second));    
109 }
110 
111 void eraseWithTreap(int v) {
112     Pair ret1 = root->split(v), ret2 = ret1.second->split(v + 1);
113     if (ret2.first->size > 1) {
114         ret2.first->snow--;
115         ret2.first->size--;
116         ret2.first->sum -= ret2.first->val;
117         root = Merge(ret1.first, Merge(ret2.first, ret2.second));
118         return;
119     }
120 
121     root = Merge(ret1.first, ret2.second);
122 }
123 
124 void insertQuery(int t) {
125     multiset <int> :: iterator it1 = s.lower_bound(t), it2 = s.upper_bound(t);
126     if (*it1 == t) {
127         s.insert(t);
128         return;
129     }
130 
131     if (it1 != s.begin()) {
132         it1--;
133     } else {
134         it1 = s.end();
135     }
136 
137     if (it1 != s.end() && it2 != s.end()) {
138         eraseWithTreap(*it2 - *it1);
139     }
140 
141     if (it1 != s.end()) {
142         insertWithTreap(t - *it1);
143     }
144 
145     if (it2 != s.end()) {
146         insertWithTreap(*it2 - t);
147     }
148 
149     s.insert(t);
150 }
151 
152 void eraseQuery(int t)
153 {
154     s.erase(s.find(t));
155     multiset <int> :: iterator it1 = s.lower_bound(t), it2 = s.upper_bound(t);
156     if (*it1 == t) {
157         return;
158     }
159 
160     if (it1 != s.begin()) {
161         it1--;
162     } else {
163         it1 = s.end();
164     }
165 
166     if (it1 != s.end() && it2 != s.end()) {
167         insertWithTreap(*it2 - *it1);
168     }
169 
170     if (it1 != s.end()) {
171         eraseWithTreap(t - *it1);
172     }
173 
174     if (it2 != s.end()) {
175         eraseWithTreap(*it2 - t);
176     }
177 }
178 
179 void askQuery(int start, int m, int r)
180 {
181     if (s.empty())
182         return;
183     ans += min((ll)(*s.begin()) * r + start, (ll)m);
184     if (r == 0) {
185         return;
186     }
187 
188     int full = m / r + ((m % r) > 0);
189     Pair ret = root->split(full);
190     ans += (ll)m * ret.second->size;
191     ans += ret.first->sum * r;
192     root = Merge(ret.first, ret.second);
193 }
194 
195 int main() {
196     null = new Node(0, 0, null, null);
197     root = null;
198     scanf("%d", &n);
199     for (int i = 1; i <= n; i++) {
200         scanf("%d %d %d", &a[i].s, &a[i].m, &a[i].r);
201     }
202 
203     scanf("%d", &m);
204     for (int i = 1; i <= m; i++) {
205         int t, l, r;
206         scanf("%d %d %d", &t, &l, &r);
207         listInsert[l].push_back(t);
208         listErase[r].push_back(t);
209     }
210 
211     for (int i = 1; i <= n; i++) {
212         for (int j = 0; j < listInsert[i].size(); j++) {
213             insertQuery(listInsert[i][j]);
214         }
215 
216         askQuery(a[i].s, a[i].m, a[i].r);
217         for (int j = 0; j < listErase[i].size(); j++) {
218             eraseQuery(listErase[i][j]);
219         }
220     }
221 
222     cout << ans << endl;    
223     return 0;
224 }

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毁灭者问题

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原文地址：http://www.cnblogs.com/AdonisGe/p/4868648.html