1635: [Usaco2007 Jan]Tallest Cow 最高的牛#
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 472 Solved: 278 [Submit][Status][Discuss]
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
差分序列大概就是前缀和的逆过程…?
这道题对于一堆关系（a,b） 要使得【a+1,b-1】中每个都严格小于a,b，但是又要尽可能大，所以令每个[a+1,b-1]每个数-1. 之前其实遇到过，以为是前缀和的第二种应用。但确切来说其实是前缀和的逆过程。。。叫差分序列？
注意去重。对于完全相同的，做一遍就好啦。所以要先排个序
1/* ***********************************************
2Author :111qqz
3Created Time :2016年04月07日 星期四 22时52分44秒
4File Name :code/bzoj/1635.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E4+7;
34int n,i,mxh,r;
35int p[N];
36
37struct node
38{
39 int a,b;
40
41 bool operator < (node y)const
42 {
43 if (a==y.a) return b<y.b;
44 return a<y.a;
45 }
46}q[N];
47int main()
48{
49 #ifndef ONLINE_JUDGE
50 freopen("code/in.txt","r",stdin);
51 #endif
52
53 ios::sync_with_stdio(false);
54 cin>>n>>i>>mxh>>r;
55 ms(p,0);
56 for ( int i = 1 ; i <= r ; i++)
57 {
58 cin>>q[i].a>>q[i].b;
59 if (q[i].a>q[i].b) swap(q[i].a,q[i].b);
60 }
61 sort(q+1,q+r+1);
62 for ( int i = 1 ;i <= r; i++)
63 {
64 int a = q[i].a;
65 int b = q[i].b;
66 if (a==q[i-1].a&&b==q[i-1].b) continue; //判重
67 p[a+1]--;
68 p[b]++;
69
70 }
71 for ( int i = 1 ; i <= n ; i++) p[i] = p[i] + p[i-1];
72 for ( int i = 1 ; i <= n ; i++) cout<<mxh+p[i]<<endl;
73 #ifndef ONLINE_JUDGE
74 fclose(stdin);
75 #endif
76 return 0;
77}