bzoj 1635: [Usaco2007 Jan]Tallest Cow 最高的牛——差分

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

————————————————————————————

这道题 差分就可以辣 a b 之间的数明显起码要比a b小1 

然后差分一下用max-前缀和就可以辣 不过这题数据有毒.... 他可以出现多组 a b

#include<cstdio>
#include<cstring>
#include<algorithm>
using std::swap;
const int M=2e4+7;
int read(){
    int ans=0,f=1,c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();}
    return ans*f;
}
int n,h,m,now;
int f[M],x,y;
struct pos{int x,y;}q[M];
bool cmp(pos a,pos b){return a.x!=b.x?a.x<b.x:a.y<b.y;}
int main(){
    n=read(); read(); h=read(); m=read();
    for(int i=1;i<=m;i++){
        x=read(); y=read();
        if(x>y) swap(x,y);
        q[i].x=x; q[i].y=y;
    }std::sort(q+1,q+1+m,cmp);
    for(int i=1;i<=m;i++){
        if(q[i].x==q[i-1].x&&q[i].y==q[i-1].y) continue;
        f[q[i].x+1]--; f[q[i].y]++;
    }
    for(int i=1;i<=n;i++){
        now+=f[i];
        printf("%d\n",h+now);
    }
    return 0;
}