Tallest Cow POJ - 3263 （区间点修改）

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.

Input

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.

Output

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

Sample Input

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

Sample Output

5
4
5
3
4
4
5
5
5


题意:n头牛 第i个牛最高，高度为h 
然后有r个关系 说明这个关系中的两头牛相互看的见（他们中间的牛高度比他们矮）
求所有牛最大可能的高度

思路：
1、既然给出每个关系中x、y两头牛相互可以看见，那么他们之间的牛的高度肯定比他要矮，所以每次给出x、y，我们只需要将x+1到y-1的牛高度全部-1，这样最后就知道了他们之间的最小高度差，
再把每个cow【i】+h   O（NR）
2、我们可以优化一下，就是说给你x、y两头牛，你在x+1的位置-1，再y的位置+1，这样我们就记录了这个关系，你从x遍历到y，让cow【i】+=cow【i-1】，你会发现，他每个x+1到y-1都是-1而x、y则是0，
我们可以把所有的关系先记录下来，然后遍历就可以知道他们之间的最小高度差，然后cow【i】+h  O（N+R）

坑点：记得去重

#include<cstdio>
#include<iostream>
#include<map>
using namespace std;
typedef pair<int,int> P;
const int maxn = 1e4+4;
int n,i,h,r;
int ans[maxn];
map<P,int>mp;
int main()
{
    scanf("%d%d%d%d",&n,&i,&h,&r);
    for(int i=1;i<=r;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        if(x > y)swap(x,y);
        if(mp[P(x,y)])continue;
        mp[P(x,y)]=1;
        for(int j=x+1;j<y;j++)
        {
            ans[j]--;
        }

    }
    for(int i=1;i<=n;i++)
    {
        printf("%d\n",ans[i]+h);
    }
}

#include<cstdio>
#include<iostream>
#include<map>
using namespace std;
typedef pair<int,int> P;
const int maxn = 1e4+4;
int n,i,h,r;
int ans[maxn];
map<P,int>mp;
int main()
{
    scanf("%d%d%d%d",&n,&i,&h,&r);
    for(int i=1;i<=r;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        if(x > y)swap(x,y);
        if(mp[P(x,y)])continue;
        mp[P(x,y)]=1;
        ans[x+1]--;
        ans[y]++;
    }
    for(int i=1;i<=n;i++)
    {
        ans[i] += ans[i-1];
        printf("%d\n",ans[i]+h);
    }
}