Gym 100886J Sockets 二分答案 + 贪心

最新推荐文章于 2019-02-21 10:51:29 发布

转载最新推荐文章于 2019-02-21 10:51:29 发布 · 100 阅读

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原文链接：http://www.cnblogs.com/liuweimingcprogram/p/5813966.html

本文介绍了一种插座分配算法，用于解决有限插座条件下多个设备供电的问题。通过二分查找和贪心策略，实现最大数量的设备同时供电。算法考虑了设备的安全级别和插座的容量。

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Description

standard input/output
Statements

Valera has only one electrical socket in his flat. He also has m devices which require electricity to work. He's got n plug multipliers to plug the devices, the i-th plug multiplier has a_i sockets.

A device or plug multiplier is supplied with electricity if it is either plugged into the electrical socket, or if it is plugged into some plug multiplier which is supplied with electricity.

For each device j, Valera knows the safety value b_j which is the maximum number of plug multipliers on the path between the device and the electrical socket in his flat. For example, if b_j = 0, the device should be directly plugged into the socket in his flat.

What is the maximum number of devices Valera could supply with electricity simultaneously? Note that all devices and plug multipliers take one socket to plug, and that he can use each socket to plug either one device or one plug multiplier.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 2·10⁵), the number of plug multipliers and the number of devices correspondingly.

The second line contains n space-separated integers a₁, a₂, ..., a_n(2 ≤ a_i ≤ 2·10⁵). Here, the number a_i stands for the number of sockets on the i-th plug multiplier.

The third line contains m space-separated integers b₁, b₂, ..., b_m(0 ≤ b_j ≤ 2·10⁵). Here, the number b_j stands for the safety value of the j-th device.

Output

Print a single integer: the maximum number of devices that could be supplied with electricity simultaneously.

坑点就是那些排插可以重复插，一个排插上可以接上多个排插。

思路：二分答案，二分出最多能插的电器，然后根据贪心，绝对是选那些要求松的，排插也是尽量选那些大的排插。

把他们都按大到小排序。

二分出一个ans，那么我就选[1,ans]这个区间的电器。

一开始，我先选择第一个排插作为基准排插，然后如果那个区间的电器有1的，就是一定要放在第一层的，就先放，如果放不下就是false，在放得下的同时，剩下的孔，应该是尽量插多一些排插，这样的结果是最优的，然后继续枚举有没有一定要放在第二层的，，一路下去即可。

特判，0是没用的。不要选。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 2e5+20;
int a[maxn],b[maxn];
int num[maxn];
int dp[maxn];
LL sum[maxn];
int n,m;
bool cmp(int a,int b)
{
    return a>b;
}
bool check (int val)
{
    int begin=1,end=val;
    LL res = a[1]; //能够插多少个
    LL can = a[1];
    int need = 1; //有没一定要插一的
    int cur = 1;
    while (1) {
        //res += a[cur];
        while (end >= begin && b[end] == need) { //一定要插这个位置
            res--; can--; end--;
            if (res < 0) return false;
        }
        if (end == begin-1) return true;
        if (can == 0) return false;
//        printf ("%d\n",n);
        while (can && cur + 1 <= n) { //插尽量多的排插
            --can; --res; ++cur;
            res += a[cur];
            //printf ("%d****\n",a[cur]);
        }
        if (cur == n) { //所有排插都用完了.
            return res >= end-begin+1;
        }
        can = res;
        need++;
       // printf ("%d\n",res);
    }
}
void work ()
{
    scanf("%d%d",&n,&m);
    for (int i=1; i<=n; ++i) scanf("%d",&a[i]);
    int lenb=0;
    for (int j=1; j<=m; ++j) {
        int x; scanf("%d",&x);
        if (x != 0) b[++lenb] = x; //0是没用的
    }
    sort (a+1,a+1+n,cmp);
    sort(b+1,b+1+lenb,cmp);

    int begin=1,end=lenb;
    while (begin<=end)
    {
        int mid = (begin+end)/2;
        if (check(mid))
            begin=mid+1;
        else end=mid-1;
    }
    printf ("%d\n",max(1,end));
    return ;
}
int main()
{
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    work();
    return 0;
}

View Code

转载于:https://www.cnblogs.com/liuweimingcprogram/p/5813966.html