Booking System
Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u
Description
Innovation technologies are on a victorious march around the planet. They integrate into all spheres of human activity!
A restaurant called “Dijkstra’s Place” has started thinking about optimizing the booking system.
There are n booking requests received by now. Each request is characterized by two numbers: ci and pi — the size of the group of visitors who will come via this request and the total sum of money they will spend in the restaurant, correspondingly.
We know that for each request, all ci people want to sit at the same table and are going to spend the whole evening in the restaurant, from the opening moment at 18:00 to the closing moment.
Unfortunately, there only are k tables in the restaurant. For each table, we know ri — the maximum number of people who can sit at it. A table can have only people from the same group sitting at it. If you cannot find a large enough table for the whole group, then all visitors leave and naturally, pay nothing.
Your task is: given the tables and the requests, decide which requests to accept and which requests to decline so that the money paid by the happy and full visitors was maximum.
Input
The first line of the input contains integer n (1 ≤ n ≤ 1000) — the number of requests from visitors. Then n lines follow. Each line contains two integers: ci, pi (1 ≤ ci, pi ≤ 1000) — the size of the group of visitors who will come by the i-th request and the total sum of money they will pay when they visit the restaurant, correspondingly.
The next line contains integer k (1 ≤ k ≤ 1000) — the number of tables in the restaurant. The last line contains k space-separated integers: r1, r2, …, rk (1 ≤ ri ≤ 1000) — the maximum number of people that can sit at each table.
Output
In the first line print two integers: m, s — the number of accepted requests and the total money you get from these requests, correspondingly.
Then print m lines — each line must contain two space-separated integers: the number of the accepted request and the number of the table to seat people who come via this request. The requests and the tables are consecutively numbered starting from 1 in the order in which they are given in the input.
If there are multiple optimal answers, print any of them.
Sample Input
Input
3
10 50
2 100
5 30
3
4 6 9
Output
2 130
2 1
3 2
Source
Codeforces Round #241 (Div. 2)
有n个订单和k个餐桌，按照某种方式选取使得总收益最大。
那么对于可以用同样的最小餐桌满足的订单，应该从金额从大到小选取，对于同样金额的订单，应该从人数从小到大选取。
那么分别排序之后贪心即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<iomanip>
#include<ctime>
#include<climits>
#include<cctype>
#include<algorithm>
#define AUTO "%I64d"
using namespace std;
#define smax(x,tmp) x=max((x),(tmp))
#define smin(x,tmp) x=min((x),(tmp))
#define maxx(x1,x2,x3) max(max(x1,x2),x3)
#define minn(x1,x2,x3) min(min(x1,x2),x3)
const int maxn = 105;
struct Edge
{
int to,next;
}edge[maxn*maxn];
int head[maxn];
int maxedge;
inline void addedge(int u,int v)
{
edge[++maxedge] = (Edge) { v,head[u] };
head[u] = maxedge;
}
int a[maxn];
int ans[maxn];
int top;
int n;
bool vis[maxn];
int main()
{
#ifndef ONLINE_JUDGE
freopen("party.in","r",stdin);
freopen("party.out","w",stdout);
#endif
scanf("%d",&n);
memset(head,-1,sizeof(head)); maxedge=-1;
char ch;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
ch = getchar();
while(!isdigit(ch)) ch=getchar();
if(ch=='1') addedge(i,j);
}
for(int i=1;i<=n;i++) scanf("%d",a+i);
while(true)
{
bool flag = false;
for(int i=1;i<=n;i++) if(!a[i])
{
ans[++top] = i;
vis[i] = true;
flag = true;
for(int j=head[i];~j;j=edge[j].next)
{
int v = edge[j].to;
a[v]--;
}
}
if(!flag) break;
}
printf("%d\n",top);
for(int i=1;i<top;i++) printf("%d ",ans[i]);
if(top) printf("%d",ans[top]);
return 0;
}
