《算法竞赛进阶指南》Tallest Cow

题目链接：http://exam.upc.edu.cn/problem.php?cid=1430&pid=6

问题 G: Tallest Cow

时间限制: 1 Sec  内存限制: 128 MB
提交: 79  解决: 34
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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.

输入

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

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

样例输入

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

样例输出

5
4
5
3
4
4
5
5
5

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 题意：

数组a[N]，给出N个数中最大的数H。R个信息x，y表示：a[x]<=a[y]，且下标为x和y之间的数严格小于a[x]。求满足条件的数组a[]

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思路：

假设初始N个数都为H，a[x+1..y-1]都减一。由于数据较小，直接暴力即可。注意给定的数据存在重复的情况。

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 代码：

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 1e4+5;

int a[maxn];
struct node
{
    int x, y;
    bool operator < (const node &a)const {
        return x < a.x;
    }
    bool operator == (const node &a)const {
        return (x==a.x && y==a.y);
    }
}node[maxn];
inline void solve(int x, int y)
{
    for(int i=x+1; i<y; ++i)
        a[i] --;
}
int main()
{

    int n, I, h, r, x, y, L, R;
    scanf("%d%d%d%d", &n, &I, &h, &r);
    for(int i=0; i<maxn; ++i) a[i]=h;

    for(int i=1; i<=r; ++i){
        scanf("%d%d", &x, &y);
        node[i].x = min(x, y);
        node[i].y = max(x, y);
    }
    sort(node+1, node+1+n);

    for(int i=1; i<=n; ++i){
        if(i>1&&node[i]==node[i-1]) continue;
        solve(node[i].x, node[i].y);
    }
    for(int i=1; i<=n; ++i) printf("%d\n", a[i]);
}