51nod 1287 Cannon two points

Source: Internet
Author: User
title: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1287 Test Instructions:

An array of positive integers of length m that represents the height of the terrain from left to right. Test a cannon, the shells parallel to the ground from left to right to fly, height is h, if the height of a terrain is greater than the height of the projectile flying H (A[i] >= h), the shells will be blocked and fall in I-1 place, then a[i-1] + 1. If H <= a[0], then this shell is invalid, if H > All A[i], this shell is not valid. Now the array B of the given n integers represents the height of the shells and calculates the final topography.
For example: terrain height a = {1, 2, 0, 4, 3, 2, 1, 5, 7}, Shell height B = {2, 8, 0, 7, 6, 5, 3, 4, 5, 6, 5}, the resulting terrain height is: {2, 2, 2, 4, 3, 3, 5, 6, 7}.

Line 1th: 2 number m, n is separated by a space, respectively, for the length of the array A and B (1 <= m, n <= 50000)
2nd to M + 1 lines: 1 numbers per line representing the corresponding terrain height (0 <= a[i] <= 1000000).
The M + 2 to n + M + 1 lines, 1 numbers per line, indicate the height of the projectile (0 <= b[i] <= 1000000).


Outputs a total of M rows, one number per line, corresponding to the final terrain height. Analysis:

As long as it is clear that the shells will only be the first one more than his >= to intercept, if the shells than AI high, then I behind the AI short can not stop. And then the problem soon falls out of the water ~

I am the second point, O (MLONGN). Look at the discussion of pre-treatment heights, and then you can do O (n). In fact, the meaning is the same. Code:

 #include <bits/stdc++.h> using namespace std; const int N = 5e4+4; int n,m,x,a[n],b[n];
    int main () {//freopen ("In.txt", "R", stdin);
    scanf ("%d%d", &n,&m);
        for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);
    B[i]=max (A[i],b[i-1]);
        } for (int i=0;i<m;i++) {scanf ("%d", &x);
        int K=lower_bound (b+1,b+1+n,x)-B; if (k>n| |
        K<=1) continue;
    if (B[k-1]<a[k-1]) b[k-1]=a[k-1];
    } for (int i=1;i<=n;i++) printf ("%d\n", A[i]);
return 0; }

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