D - Supermarket

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Supermarket

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Time Limit: 10 Seconds       Memory Limit: 32768 KB

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A supermarket has a set Prod of products on sale. It earns a profit px for each product x in Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell (subset of Prod) such that the selling of each product x in Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=sum of px (x in Sell). An optimal selling schedule is a schedule with a maximum profit.

For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

schedule	profit
{a}	50
{b}	10
{c}	20
{d}	30
{b,a}	60
{a,c}	70
{c,a}	70
{b,c}	30
{d,a}	80
{d,c}	50

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

The sample input in contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.

Sample Input

4 50 2 10 1 20 2 30 1

7 20 1 2 1 10 3 100 2 8 2
5 20 50 10

Sample Output

80
185

题意：n个数，后面n个例子是 东西的利润和卖的截止日期。 东西得在截止日期之前卖才能有利润。问利润最大是多少。

//china no.1
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define rand() srand(time(0));
#define me(x) memset(x,0,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,-1,-1,1,1};
const int dy[]={0,1,0,-1,1,-1,1,-1};
const int maxn=1e3+100;
const int maxx=1e4+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;
long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
inline int Scan()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')flag=1;
    else if(ch>='0' && ch<='9')res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';
    return flag ? -res : res;
}

int n;
struct node
{
    LL v,day;
}Q[maxx];
bool cmp(node a,node b)
{
    if(a.v==b.v) return a.day<b.day;
    return a.v>b.v;
}
int vis[maxx];
int main()
{
    //close();
    W(scanf("%d",&n)!=EOF)
    {
        me(Q);
        me(vis);
        FOR(1,n,i)
            cin>>Q[i].v>>Q[i].day;
        sort(Q+1,Q+n+1,cmp);
        int ans=0;
        FOR(1,n,i)
        {
            for(int j=Q[i].day;j>=1;j--)
                if(!vis[j])
            {
                vis[j]=1;
                ans+=Q[i].v;
                break;
            }
        }
        cout<<ans<<endl;
    }
}