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

Sample Output

5 4 5 3 4 4 5 5 5

思路：用差分?jǐn)?shù)組維護(hù)區(qū)間的增減，坑點是有可能出現(xiàn)相同區(qū)間，如果出現(xiàn)相同的區(qū)間就不用處理了，真坑，wa了，用map標(biāo)記之前是否出現(xiàn)過

其余應(yīng)該比較簡單

代碼：

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<vector> #include<cmath> #include<set> #include<stack> #include<map> #define MAX 10005typedef long long ll;using namespace std;map<int,int>mp; int a[MAX]; int d[MAX]; int vis[1000005]; int main() {int N,I,H,R;cin>>N>>I>>H>>R;for(int t=1;t<=N;t++){a[t]=H;}for(int t=1;t<=N;t++){d[t]=a[t]-a[t-1];}int l,r;for(int t=0;t<R;t++){scanf("%d%d",&l,&r);if(max(l,r)-min(l,r)>=2&&mp[l]!=r){d[min(l,r)+1]--;d[max(l,r)]++;mp[l]=r;}}for(int t=1;t<=N;t++){a[t]=a[t-1]+d[t];}for(int t=1;t<=N;t++){cout<<a[t]<<endl;}return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/Staceyacm/p/10781798.html

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