POJ-3263 Tallest Cow

Tallest Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5425		Accepted: 2500

Description

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

Source

USACO 2007 January Silver

ID-OJ:
POJ-3263

author:
Caution_X

DATE of submission:
20,191,117

Tags:
difference, prefix and

description modelling:
n-cows in a row, and the highest known reference height H and R cattle relational
relationship: xi and yi two cows can see each other, i.e., between xi and yi bovine cattle than they short
the output of the maximum height of each cow

Major Steps to Solve IT:
(. 1) each with an array of labeled xi and yi, a [xi + 1 ] = + -. 1, a [Yi] + =. 1
(2) using tmp prefix and stores the current, the current maximum height bovine tmp + H =

the AC code:

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
using namespace std;
typedef pair<int,int> P;
set<P> book;
int a[10005];
int main()
{
    //freopen("input.txt","r",stdin);
    int N,I,H,R;
    scanf("%d%d%d%d",&N,&I,&H,&R);
    for(int i=0;i<R;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        int temp=min(x,y);
        y=max(x,y);
        x=temp;
        P tmp=P(x,y);
        if(book.find(tmp)==book.end()&&abs(x-y)>1)
        {
            a[x+1]--;
            a[y]++;
            book.insert(tmp);
        }
    }
    int tmp=0;
    for(int i=1;i<=N;i++)
    {    
        tmp+=a[i];
        //cout<<tmp<<' ';
        printf("%d\n",tmp+H);
    }
    return 0;
}