牛客算法周周练3/POJ - 3614 E -Sunscreen（贪心，优先队列）

题目描述

To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi ≤ maxSPFi ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn't tan at all........
The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri cows with lotion. A cow may lotion from only one bottle.
What is the maximum number of cows that can protect themselves while tanning given the available lotions?

输入描述:

* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i's lotion requires with two integers: minSPFi and maxSPFi
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri

输出描述:

A single line with an integer that is the maximum number of cows that can be protected while tanning

示例1

输入

复制

输出

复制

链接：https://ac.nowcoder.com/acm/contest/5338/E
来源：牛客网

题意：

奶牛美容：有 $\mathit C$ 头奶牛日光浴，每头奶牛分别需要 $minSPF_i$ 和 $maxSPF_i$ 单位强度之间的阳光。现有 $\mathit L$ 种防晒霜，分别能使阳光强度稳定为 $SPF_i$ ，其瓶数为 $cover_i$ 。求最多满足多少头奶牛?

思路：

将奶牛按照 $minSPF_i$ 为第一指标， $maxSPF_i$ 为第二指标进行升序排序。

将防晒霜按照 $SPF_i$ 指标进行升序排序，

从 $1->m$ 枚举第 $\mathit i$ 个防晒霜，将满足 $minSPF_j\leq SPF_i$ 的奶牛的 $maxSPF_j$ 都加入到一个小根堆中，

然后对于堆顶的元素如果满足： $maxSPF_j >= SPF_i$ 就将一个第 $\mathit i$ 个防晒霜给这个奶牛使用。

核心思想：对于满足 $minSPF_j\leq SPF_i$ 的奶牛，优先第 $\mathit i$ 个防晒霜给 $maxSPF_j$ 较小的使用，

因为 $maxSPF_j$ 较大的有较大的选择空间。

可以思考下面样例，方便理解：

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <unordered_map>
// #include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x)  if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n, m;
pii a[maxn];
pii b[maxn];

bool cmp2(pii & aa, pii & bb)
{
    return aa.fi < bb.fi;
}
priority_queue<int, vector<int>, greater<int> > heap;
int main()
{
#if DEBUG_Switch
    freopen("C:\\code\\input.txt", "r", stdin);
#endif
    //freopen("C:\\code\\output.txt","r",stdin);
    n = readint();
    m = readint();
    repd(i, 1, n)
    {
        a[i].fi = readint();
        a[i].se = readint();
    }
    sort(a + 1, a + 1 + n);
    repd(i, 1, m)
    {
        b[i].fi = readint();
        b[i].se = readint();
    }
    sort(b + 1, b + 1 + m, cmp2);
    int id = 1;
    int ans = 0;
    repd(i, 1, m)
    {
        while (id <= n && a[id].fi <= b[i].fi)
        {
            heap.push(a[id].se);
            id++;
        }
        while (!heap.empty() && b[i].se > 0)
        {
            int x = heap.top();
            heap.pop();
            if (x >= b[i].fi)
            {
                b[i].se--;
                ans++;
            } else
            {
                continue;
            }
        }
    }
    printf("%d\n", ans );
    return 0;
}