[题解] POJ-3263 Tallest Cow

Tallest Cow

Time Limit: 2000MS
Memory Limit: 65536K

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

解题思路

题目要使每头牛的身高尽可能高,并且给出了 $M$ 对关系,$A_i$ 和 $B_i$ 可以互相看见,这也代表了 $[A_i+1, B_i-1] $ 每头牛的身高都比 $A_i$ 和 $B_i$ 低。

考虑用一个数组维护这个关系,将$[A_i+1, B_i-1]$ 对应的值减去1,复杂度为 $O(MN)$。我们可以将 $A_i+1$ 减去1,$B_i+1$ 加上1,最后求出他们的前缀和。

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
const int MAXN = 10000 + 10;

inline int read(){
    int x = 0;
    char ch = getchar();

    while(ch < '0' || ch > '9'){
        ch = getchar();
    }

    while('0' <= ch && ch <= '9'){
        x = x*10 + ch - '0';
        ch = getchar();
    }

    return x;
} 

map , bool> exists;

int data[MAXN];
int N, I, H, R;

int main(){
    N = read();
    I = read();
    H = read();
    R = read();

    for(register int i=0; i b)
            swap(a, b);

        if(exists[make_pair(a,b)])
            continue;

        data[a+1]--;
        data[b]++;

        exists[make_pair(a,b)] = true;

    }

    for(register int i=1; i<=N; i++){
        data[i] += data[i-1];

        printf("%dn", data[i] + H);
    }   
    return 0;
}