【河南省多校脸萌第六场 B】点兵点将

【链接】点击打开链接


【题意】


在这里写题意

【题解】


先每个单位都不建造bi;
打死一个ai之后,把bi加入到大根堆里面.
然后等到不够打死某个单位的时候;
从大根堆里面取出最大的那个bi;不断取,直到够打死ai为止。

【错的次数】


0

【反思】


m写成了n

【代码】

/*
 
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <cstdlib>
#include <cmath>
#include <bitset>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb emplace_back
#define fi first
#define se second
#define ld long double
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define rf(x) scnaf("%lf",&x)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
#define sz(x) ((int) x.size())
#define ld long double
 
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
 
//mt19937 myrand(time(0));
//int get_rand(int n){return myrand()%n + 1;}
const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 5e4;
 
int n, m;
priority_queue <int, vector<int>, less <int> > q2;
int a[N + 10], b[N + 10],rest;
 
int getans() {
    int ans = 0;
    rest = n;
    rep1(i, 1, m) {
        if (rest >= a[i]) {
            rest -= a[i];
            q2.push(b[i]);
        }
        else {
            a[i] -= rest;
            while (1) {
                if (q2.empty()) return -1;
                rest = q2.top();
                q2.pop();
                ans++;
                if (rest >= a[i]) {
                    rest -= a[i];
                    a[i] = 0;
                    break;
                }
                else
                    a[i] -= rest;
            }
            q2.push(b[i]);
        }
    }
    return ans;
}
 
int main() {
    //Open();
    //Close();
    while (~ri(n)) {
        ri(m);
        rep1(i, 1, m) ri(a[i]);
        rep1(i, 1, m) ri(b[i]);
        while (!q2.empty()) q2.pop();
        oi(getans()); puts("");
    }
    return 0;
}


原文地址:https://www.cnblogs.com/AWCXV/p/7626059.html