1635: [Usaco2007 Jan]Tallest Cow 最高的牛

Time Limit: 5 Sec Memory Limit: 64 MB

Submit: 383 Solved: 211

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

HINT

Source

Silver

题解：好吧我承认我真心捉鸡（phile：又逗比了？ HansBug：才没哪，讨厌啦）——我以前一直在用一种数组，这个数组的前缀和实际上表示的才是需要的值，这个东西在写树状数组时频频用到，直到今天我才知道这个是差分序列（QAQ）。。。然后这个没啥啦。。。直接代码——

 1 var
 2    i,j,k,l,m,n,h,r:longint;
 3    a:array[0..20000,1..2] of longint;
 4    b:array[0..20000] of longint;
 5 procedure swap(var x,y:longint);
 6           var z:longint;
 7           begin
 8                z:=x;x:=y;y:=z;
 9           end;
10 
11 procedure sort(l,r,z:longint);
12           var i,j,x,y:longint;
13           begin
14                i:=l;j:=r;x:=a[(l+r) div 2,z];
15                repeat
16                      while a[i,z]<x do inc(i);
17                      while a[j,z]>x do dec(j);
18                      if i<=j then
19                         begin
20                              swap(a[i,1],a[j,1]);
21                              swap(a[i,2],a[j,2]);
22                              inc(i);dec(j);
23                         end;
24                until i>j;
25                if l<j then sort(l,j,z);
26                if i<r then sort(i,r,z);
27           end;
28 begin
29      readln(n,i,h,r);
30      for i:=1 to r do
31          begin
32               readln(a[i,1],a[i,2]);
33               if a[i,1]>a[i,2] then swap(a[i,1],a[i,2]);
34          end;
35      sort(1,r,1);
36      j:=1;
37      for i:=1 to r+1 do
38          begin
39               if a[i,1]<>a[j,1] then
40                  begin
41                       sort(j,i-1,2);
42                       j:=i;
43                  end;
44          end;
45      for i:=1 to r do
46          begin
47               if (a[i,1]=a[i-1,1]) and (a[i,2]=a[i-1,2]) then continue;
48               inc(b[a[i,1]+1]);dec(b[a[i,2]]);
49          end;
50      l:=0;
51      for i:=1 to n do
52          begin
53               l:=l+b[i];
54               writeln(-l+h);
55          end;
56 end.