hdu 4864

Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1930    Accepted Submission(s): 496

Problem Description

Today the company has m tasks to complete. The ith task need xi minutes to complete. Meanwhile, this task has a difficulty level yi. The machine whose level below this task’s level yi cannot complete this task. If the company completes this task, they will get (500*xi+2*yi) dollars.
The company has n machines. Each machine has a maximum working time and a level. If the time for the task is more than the maximum working time of the machine, the machine can not complete this task. Each machine can only complete a task one day. Each task can only be completed by one machine.
The company hopes to maximize the number of the tasks which they can complete today. If there are multiple solutions, they hopes to make the money maximum.

 

Input

The input contains several test cases. 
The first line contains two integers N and M. N is the number of the machines.M is the number of tasks(1 < =N <= 100000,1<=M<=100000).
The following N lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the maximum time the machine can work.yi is the level of the machine.
The following M lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the time we need to complete the task.yi is the level of the task.

 

Output

For each test case, output two integers, the maximum number of the tasks which the company can complete today and the money they will get.

 

Sample Input

1 2100 3100 2100 1

 

Sample Output

1 50004

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

基本思想是贪心。

对于价值c=500*xi+2*yi，yi最大影响100*2<500,所以就是求xi总和最大。可以先对机器和任务的时间从大到小排序。从最大时间的任务开始，找出满足任务时间要求的所有机器，从中找出等级最低且满足任务等级要求的机器匹配。依次对任务寻找满足要求的机器。

用multiset可以处理相同值的情况，erase（）的参数可以是迭代器，也可以是元素值

#include<stdio.h>#include<algorithm>#include<set>#define N 100005using namespace std;multiset<int>g;struct node{    int x,y;}a[N],b[N];bool cmp(node a,node b){    if(a.x!=b.x)        return a.x>b.x;    else        return a.y>b.y;}int main(){    //freopen("1004.in","r",stdin);    //freopen("1004.out","w",stdout);    int n,m,i,j,num;    long long sum;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=0;i<n;i++)            scanf("%d%d",&a[i].x,&a[i].y);        for(i=0;i<m;i++)            scanf("%d%d",&b[i].x,&b[i].y);        sort(b,b+m,cmp);        sort(a,a+n,cmp);        j=0;num=0,sum=0;        g.clear();        for(i=0;i<m;i++)        {            while(j<n&&a[j].x>=b[i].x)            {                g.insert(a[j].y);                j++;            }           multiset<int>::iterator it=g.lower_bound(b[i].y);           if(it!=g.end())           {               num++;               sum+=500*b[i].x+2*b[i].y;               g.erase(it);           }        }        printf("%d %I64d\n",num,sum);    }    return 0;}

#include<stdio.h>#include<algorithm>#include<set>#define N 100005using namespace std;multiset<int>g[105];struct node{    int x,y;}b[N];bool cmp(node a,node b){    if(a.x!=b.x)        return a.x>b.x;    else        return a.y>b.y;}int main(){    //freopen("1004.in","r",stdin);    //freopen("1004.out","w",stdout);    int n,m,i,j,num,x,y;    long long sum;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=0;i<=100;i++)            g[i].clear();        for(i=0;i<n;i++)        {         scanf("%d%d",&x,&y);         g[y].insert(x);        }        for(i=0;i<m;i++)            scanf("%d%d",&b[i].x,&b[i].y);        sort(b,b+m,cmp);        num=0,sum=0;        for(i=0;i<m;i++)        {            x=b[i].x;            y=b[i].y;            for(j=y;j<=100;j++)            {                if(g[j].empty()==true)                    continue;               multiset<int>::iterator it=g[j].lower_bound(x);               if(it!=g[j].end())               {                   num++;                   sum+=500*b[i].x+2*b[i].y;                   g[j].erase(it);//删除当前迭代器指向的元素                   break;               }            }        }        printf("%d %I64d\n",num,sum);    }    return 0;}