51Nod 1279 扔盘子 (思维+模拟)

题意:

有口井,往里扔盘子,最多扔多少个

n<=5e5, 1s

思路:

如果比较高的地方井口比较小,那么下面的再大也没有用,只需要维护一个单调减的数组然后O(n+m)模拟即可

代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);


int a[maxn];
int b[maxn];
int n,m;
int main(){
    
    int tmp = inf;
    scanf("%d %d", &n,&m);
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        tmp = min(tmp, a[i]);
        a[i] = min(tmp, a[i]);
    }
    int ans = 0;
    int p = n;
    for(int i = 1; i <= m; i++){
        int x;
        scanf("%d", &x);
        while(p>=1&&a[p]<x){
            p--;
        }
        if(p){p--;ans++;}

    }
    printf("%d",ans);
    return 0;

}

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转载自www.cnblogs.com/wrjlinkkkkkk/p/10544000.html