[Luogu2879][USACO07JAN]区间统计Tallest Cow

题目描述

FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow.

FJ has made a list of R (0 ≤ R ≤ 10,000) lines of the form "cow 17 sees cow 34". This means that cow 34 is at least as tall as cow 17, and that every cow between 17 and 34 has a height that is strictly smaller than that of cow 17.

For each cow from 1..N, determine its maximum possible height, such that all of the information given is still correct. It is guaranteed that it is possible to satisfy all the constraints.

给出牛的可能最高身高，然后输入R组数据 a b,代表a，b可以相望，最后求所有牛的可能最高身高输出

输入输出格式

输入格式：

Line 1: Four space-separated integers: N, I, H and R

Lines 2..R+1: Two distinct space-separated integers A and B (1 ≤ A, B ≤ N), indicating that cow A can see cow B.

输出格式：

Lines 1..N: Line i contains the maximum possible height of cow i.

输入输出样例

输入样例#1：

9 3 5 5
1 3
5 3
4 3
3 7
9 8

输出样例#1：

5
4
5
3
4
4
5
5
5

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我们对于每一个输入(x, y)，把x+1~y区间减一。
这个模拟一下应该就看出来了，然后用树状数组维护差分序列。
本来想着一发走人，然而只有50分...ri...
这题丧心病狂有重复输入，不判重一半分没有...
好像没必要用树状数组，直接用差分数组，反正也要从头扫一遍...
emm...貌似比我的复杂度优秀，管他呢能过就行...

---

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
inline int read(){
    int res=;char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch)){res=(res<<)+(res<<)+(ch^);ch=getchar();}
    return res;
}
int n, H, I, Q;
int tr[];
#define lowbit(x) x&-x
inline void add(int x, int y)
{
    while(x <= n)
    {
        tr[x] += y;
        x += lowbit(x);
    }
}
inline int ask(int x)
{
    int res = ;
    while(x)
    {
        res += tr[x];
        x -= lowbit(x);
    }
    return res;
}
map <pair<int,int>,int>mp;

int main()
{
    n = read(), I = read(), H = read(), Q = read();
    while(Q--)
    {
        int x = read(), y = read();
        if (mp[make_pair(x, y)]) continue;
        if(x > y)swap(x, y);
        add(x+, -);
        add(y, );
        mp[make_pair(x, y)] = ;
        mp[make_pair(y, x)] = ;
    }
    for (int i =  ; i <= n ; i ++)
    {
        printf("%d\n",H + ask(i));
    }
    return ;
}