zhber
有好多做过的题没写下来,如果我还能记得就补吧

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.

有n(1 <= n <= 10000)头牛从1到n线性排列,每头牛的高度为h[i](1 <= i <= n),现在告诉你这里面的牛的最大高度为maxH,而且有r组关系,每组关系输入两个数字,假设为a和b,表示第a头牛能看到第b头牛,能看到的条件是a, b之间的其它牛的高度都严格小于min(h[a], h[b]),而h[b] >= h[a]

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


INPUT DETAILS:

There are 9 cows, and the 3rd is the tallest with height 5.

Sample Output

5
4
5
3
4
4
5
5
5

这题要用差分序列来稿……

差分序列是个好东西……O(m)的效率支持区间加减,最后可以用前缀和搞区间询问。如果要在线询问的话要加个树状数组

对于题目的每个关系(a,b),只要f[a+1]--、f[b]++就好了

不过这题有点坑,询问会重。所以还要sort判重

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
struct work{
	int l,r;
}dat[20001];
inline bool cmp(const work &a,const work &b)
{return a.l<b.l||a.l==b.l&&a.r<b.r;}
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,useless,h,r,x,y,sum;
int chafen[100001];
int main()
{
	n=read();
	useless=read();
	h=read();
	r=read();
	for (int i=1;i<=r;i++)
	  {
		dat[i].l=read();
		dat[i].r=read();
		if (dat[i].r<dat[i].l)swap(dat[i].l,dat[i].r);
	  }
	sort(dat+1,dat+r+1,cmp);
	for (int i=1;i<=r;i++)
	  {
	  	if (dat[i].l==dat[i-1].l&&dat[i].r==dat[i-1].r)continue;
	  	chafen[dat[i].l+1]++;
	  	chafen[dat[i].r]--;
	  }
	for (int i=1;i<=n;i++)
	  {
	  	sum+=chafen[i];
	  	printf("%d\n",h-sum);
	  }
	return 0;
}


posted on 2014-07-29 13:39  zhber  阅读(112)  评论(0编辑  收藏  举报