# BZOJ1635: [Usaco2007 Jan]Tallest Cow 最高的牛

## 1635: [Usaco2007 Jan]Tallest Cow 最高的牛

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 346  Solved: 184
[Submit][Status]

## 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.

## 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.

5
4
5
3
4
4
5
5
5

## Source

Silver  1 #include<cstdio>
2
3 #include<cstdlib>
4
5 #include<cmath>
6
7 #include<cstring>
8
9 #include<algorithm>
10
11 #include<iostream>
12
13 #include<vector>
14
15 #include<map>
16
17 #include<set>
18
19 #include<queue>
20
21 #include<string>
22
23 #define inf 1000000000
24
25 #define maxn 10000+500
26
27 #define maxm 500+100
28
29 #define eps 1e-10
30
31 #define ll long long
32
33 #define pa pair<int,int>
34
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42
43 #define mod 1000000007
44
45 using namespace std;
46
47 inline int read()
48
49 {
50
51     int x=0,f=1;char ch=getchar();
52
53     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
54
55     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
56
57     return x*f;
58
59 }
60 int b[maxn],n,ans,m;
61 struct rec{int l,r;}a[maxn];
62 inline bool cmp(rec a,rec b)
63 {
64     return a.l<b.l||(a.l==b.l&&a.r<b.r);
65 }
66
67 int main()
68
69 {
70
71     freopen("input2.txt","r",stdin);
72
73     freopen("output3.txt","w",stdout);
74
75     n=read();ans=read();ans=read();m=read();
76     for1(i,m)
77      {
78          a[i].l=read();a[i].r=read();
79          if(a[i].l>a[i].r)swap(a[i].l,a[i].r);
80      }
81     sort(a+1,a+m+1,cmp);
82     for1(i,m)
83     {
84         if(a[i].l==a[i-1].l&&a[i].r==a[i-1].r)continue;
85         b[a[i].l+1]--;b[a[i].r]++;
86     }
87     for1(i,n)
88     {
89          ans+=b[i];
90         printf("%d\n",ans);
91     }
92
93     return 0;
94
95 }
View Code

posted @ 2014-09-19 13:18  ZYF-ZYF  Views(145)  Comments(0Edit  收藏  举报