Description
Byteotian Interstellar Union (BIU) has recently discovered a new planet in a nearby galaxy. The planet is unsuitable for colonisation due to strange meteor showers, which on the other hand make it an exceptionally interesting object of study.
The member states of BIU have already placed space stations close to the planet‘s orbit. The stations‘ goal is to take samples of the rocks flying by. The BIU Commission has partitioned the orbit into msectors, numbered from 1to m, where the sectors 1and mare adjacent. In each sector there is a single space station, belonging to one of the nmember states.
Each state has declared a number of meteor samples it intends to gather before the mission ends. Your task is to determine, for each state, when it can stop taking samples, based on the meter shower predictions for the years to come.
Byteotian Interstellar Union有N个成员国。现在它发现了一颗新的星球，这颗星球的轨道被分为M份（第M份和第1份相邻），第i份上有第Ai个国家的太空站。
这个星球经常会下陨石雨。BIU已经预测了接下来K场陨石雨的情况。 BIU的第i个成员国希望能够收集Pi单位的陨石样本。你的任务是判断对于每个国家，它需要在第几次陨石雨之后，才能收集足够的陨石。 输入： 第一行是两个数N,M。 第二行有M个数，第i个数Oi表示第i段轨道上有第Oi个国家的太空站。 第三行有N个数，第i个数Pi表示第i个国家希望收集的陨石数量。 第四行有一个数K，表示BIU预测了接下来的K场陨石雨。 接下来K行，每行有三个数Li,Ri,Ai，表示第K场陨石雨的发生地点在从Li顺时针到Ri的区间中（如果Li<=Ri，就是Li,Li+1,...,Ri，否则就是Ri,Ri+1,...,m-1,m,1,...,Li），向区间中的每个太空站提供Ai单位的陨石样本。 输出： N行。第i行的数Wi表示第i个国家在第Wi波陨石雨之后能够收集到足够的陨石样本。如果到第K波结束后仍然收集不到，输出NIE。 数据范围：
数据范围： 1<=n,m,k<=3*10^5 1<=Pi<=10^9 1<=Ai<10^9
Input
The first line of the standard input gives two integers, n and m(1<=n,m<=3*10^5) separated by a single space, that denote, respectively, the number of BIU member states and the number of sectors the orbit has been partitioned into.
In the second line there are mintegers Oi(1<=Oi<=n) separated by single spaces, that denote the states owning stations in successive sectors.
In the third line there are nintegers Pi(1<=Pi<=10^9) separated by single spaces, that denote the numbers of meteor samples that the successive states intend to gather.
In the fourth line there is a single integer k(1<=k<=3*10^5) that denotes the number of meteor showers predictions. The following klines specify the (predicted) meteor showers chronologically. The i-th of these lines holds three integers Li, Ri, Ai(separated by single spaces), which denote that a meteor shower is expected in sectors Li,Li+1,…Ri (if Li<=Ri) or sectors Li,Li+1,…,m,1,…Ri (if Li>Ri), which should provide each station in those sectors with Aimeteor samples (1<=Ai<10^9).
In tests worth at least 20% of the points it additionally holds that .
Output
Your program should print nlines on the standard output. The i-th of them should contain a single integer Wi, denoting the number of shower after which the stations belonging to the i-th state are expected to gather at least Pi samples, or the word NIE (Polish for no) if that state is not expected to gather enough samples in the foreseeable future.
Sample Input
3 5
1 3 2 1 3
10 5 7
3
4 2 4
1 3 1
3 5 2
Sample Output
3
NIE
1
HINT
Source
因为之前用可持久化线段树被卡内存了，表示不爽，于是用整体二分来"水掉"这道题。
整体二分其实并没有什么特别高端的东西。
单次询问二分答案使得总时间复杂度太高了，所以就把所有操作和询问放到一起进行二分答案。用一个数组来维护答案在当前二分的区间[l, r]内的询问。在每一层中"暴力"进行用数据结构进行计算1 ~ mid的操作对答案的贡献，如果比需求的多或者等于，就扔进左区间的队列，然后递归处理，否则扔进右区间的队列递归处理。
同时注意保障每一层的时间复杂度，最坏的情况下，"解答树"是一棵满二叉树，如果真的是暴力执行 1 ~ mid 的操作和直接二分答案就没什么区别了(所以说要巧妙地利用已有的计算结果)。
对于这道题，这个数据结构就用树状数组就好了(没有必要出动线段树这种大牛数据结构)
Code
1 /**
2 * bzoj
3 * Problem#2527
4 * Accepted
5 * Time:13520ms
6 * Memory:16336k
7 */
8 #include <iostream>
9 #include <cstdio>
10 #include <ctime>
11 #include <cmath>
12 #include <cctype>
13 #include <cstring>
14 #include <cstdlib>
15 #include <fstream>
16 #include <sstream>
17 #include <algorithm>
18 #include <map>
19 #include <set>
20 #include <stack>
21 #include <queue>
22 #include <vector>
23 #ifndef WIN32
24 #define Auto "%lld"
25 #else
26 #define Auto "%I64d"
27 #endif
28 using namespace std;
29 typedef bool boolean;
30 const signed int inf = (signed)((1u << 31) - 1);
31 const signed long long llf = (signed long long)((1ull << 63) - 1);
32 const double eps = 1e-6;
33 const int binary_limit = 128;
34 #define smin(a, b) a = min(a, b)
35 #define smax(a, b) a = max(a, b)
36 #define max3(a, b, c) max(a, max(b, c))
37 #define min3(a, b, c) min(a, min(b, c))
38 template<typename T>
39 inline boolean readInteger(T& u){
40 char x;
41 int aFlag = 1;
42 while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
43 if(x == -1) {
44 ungetc(x, stdin);
45 return false;
46 }
47 if(x == ‘-‘){
48 x = getchar();
49 aFlag = -1;
50 }
51 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u * 10) + x - ‘0‘);
52 ungetc(x, stdin);
53 u *= aFlag;
54 return true;
55 }
56
57 #define LL long long
58 #define lowbit(x) (x & (-x))
59
60 typedef class IndexedTree {
61 public:
62 LL* a;
63 int local;
64 int s;
65 IndexedTree():a(NULL), s(0), local(0) { }
66 IndexedTree(int n):s(n), local(0) {
67 a = new LL[(n + 2)];
68 memset(a, 0, sizeof(LL) * (n + 2));
69 }
70
71 inline void add(int idx, LL val) {
72 for(; idx <= s; idx += lowbit(idx))
73 a[idx] += val;
74 }
75
76 inline void add(int l, int r, LL val) {
77 add(l, val);
78 add(r + 1, -val);
79 }
80
81 inline LL getSum(int idx) {
82 LL rt = 0;
83 for(; idx; idx -= lowbit(idx))
84 rt += a[idx];
85 return rt;
86 }
87 }IndexedTree;
88
89 typedef class Updater {
90 public:
91 int l;
92 int r;
93 int val;
94 }Updater;
95
96 int n, m, q;
97 vector<int> *owns;
98 int* goals;
99 Updater *us;
100
101 inline void init() {
102 readInteger(n);
103 readInteger(m);
104 owns = new vector<int>[(n + 1)];
105 goals = new int[(n + 1)];
106 for(int i = 1, x; i <= m; i++) {
107 readInteger(x);
108 owns[x].push_back(i);
109 }
110 for(int i = 1; i <= n; i++)
111 readInteger(goals[i]);
112 readInteger(q);
113 us = new Updater[(q + 2)];
114 for(int i = 1; i <= q; i++) {
115 readInteger(us[i].l);
116 readInteger(us[i].r);
117 readInteger(us[i].val);
118 }
119 us[q + 1] = (Updater) {1, 1, 0};
120 }
121
122 inline void modify(IndexedTree& it, int i, int sign) {
123 if(us[i].l > us[i].r)
124 it.add(us[i].l, m, us[i].val * sign), it.add(1, us[i].r, us[i].val * sign);
125 else
126 it.add(us[i].l, us[i].r, us[i].val * sign);
127 }
128
129 int* res;
130 IndexedTree it;
131 void CDQDividing(queue<int> &que, int l, int r) {
132 if(que.empty()) return;
133 if(l == r) {
134 while(!que.empty()) {
135 int x = que.front();
136 que.pop();
137 res[x] = l;
138 }
139 return;
140 }
141
142 int mid = (l + r) >> 1;
143 while(it.local < mid)
144 modify(it, ++it.local, 1);//, cntit++;
145 while(it.local > mid)
146 modify(it, it.local--, -1);//, cntit++;
147
148 // fprintf(stderr, "dep-%d time %dms %d", dep, clock(), cntit);
149
150 queue<int> ql, qr;
151 while(!que.empty()) {
152 int i = que.front();
153 que.pop();
154 LL s = 0;
155 for(int j = 0; j < (signed)owns[i].size() && s <= goals[i]; j++)
156 s += it.getSum(owns[i][j]);
157 if(s < goals[i])
158 qr.push(i);
159 else
160 ql.push(i);
161 }
162
163 // while(!que.empty()) que.pop();
164 // memset(it.a, 0, sizeof(LL) * (m + 2));
165 CDQDividing(ql, l, mid);
166 CDQDividing(qr, mid + 1, r);
167 }
168
169 queue<int> que;
170 inline void solve() {
171 res = new int[(n + 1)];
172 for(int i = 1; i <= n; i++)
173 que.push(i);
174 it = IndexedTree(m + 1);
175 CDQDividing(que, 1, q + 1);
176 for(int i = 1; i <= n; i++)
177 if(res[i] <= q)
178 printf("%d\n", res[i]);
179 else
180 puts("NIE");
181 }
182
183 int main() {
184 // freopen("meteors.in", "r", stdin);
185 // freopen("meteors.out", "w", stdout);
186 init();
187 solve();
188 return 0;
189 }