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)頭牛從１到n線性排列，每頭牛的高度為h[i](1 <= i <= n),現在告訴你這里面的牛的最大高度為maxH,而且有r組關系，每組關系輸入兩個數字，假設為a和b,表示第a頭牛能看到第b頭牛，能看到的條件是a, b之間的其它牛的高度都嚴格小于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
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
4
5
3
4
4
5
5
5
#include<cstdio>
#include<cstring>
#include<algorithm>
using std::swap;
const int M=2e4+;
int read(){
int ans=,f=,c=getchar();
while(c<''||c>''){if(c=='-') f=-; c=getchar();}
while(c>=''&&c<=''){ans=ans*+(c-''); 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=;i<=m;i++){
x=read(); y=read();
if(x>y) swap(x,y);
q[i].x=x; q[i].y=y;
}std::sort(q+,q++m,cmp);
for(int i=;i<=m;i++){
if(q[i].x==q[i-].x&&q[i].y==q[i-].y) continue;
f[q[i].x+]--; f[q[i].y]++;
}
for(int i=;i<=n;i++){
now+=f[i];
printf("%d\n",h+now);
}
return ;
}
總結
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