1635: [Usaco2007 Jan]Tallest Cow 最高的牛
Time Limit: 5 Sec
Memory Limit: 64 MB
Submit: 438
Solved: 253
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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
一开始居然想用拓扑。。。
首先能证明没有两个区间相交，那么最多端点重合，然后去重差分就好。。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<cstdlib>
#include<map>
#include<cmath>
using namespace std;
const int maxn = 1e4 + 10;
int n,i,j,h[maxn],T,r,I;
vector <int> v[maxn];
int main()
{
#ifndef ONLINE_JUDGE
#ifndef YZY
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#else
freopen("yzy.txt","r",stdin);
#endif
#endif
cin >> n >> I >> T >> r;
while (r--)
{
int L,R;
scanf("%d%d",&L,&R);
int from = min(L,R);
int to = max(L,R);
bool have = false;
for (i = 0; i < v[from].size(); i++)
if (v[from][i] == to)
{
have = true;
break;
}
if (have) continue;
v[from].push_back(to);
}
for (i = 1; i <= n; i++)
for (j = 0; j < v[i].size(); j++)
--h[i + 1],++h[v[i][j]];
for (i = 1; i <= n; i++) h[i] += h[i - 1],printf("%d\n",h[i] + T);
return 0;
}
