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.

输入描述:

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.

输出描述:

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

输入样例:

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

输出样例:

5
4
5
3
4
4
5
5
5

题目分析

此题简单,构造差分数组再前缀和还原即可。 差分数组中,最高的牛对于的值一定为 0, 所以最后前缀和还原后的高度需要加上偏移量H

代码实现

#include <iostream>
#include <bitset>
using namespace std;
const int N = 1e4 + 10;

int n, i, h, r, cow[N], a, b;
bitset <N> bt[N];
int main () {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> i >> h >> r;
    while(r--) {
        cin >> a >> b;
        if(a > b) swap(a, b);
        if(!bt[a][b]) {
            --cow[a + 1], ++cow[b];    
            bt[a][b] = 1;
        }
    }
    for(int i = 1; i <= n; ++i) {
        cow[i] += cow[i - 1];
        cout << cow[i] + h << '\n';
    }
    return 0;
}

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