Poj1827a bunch of monsters (Greedy)

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Give n monsters the range of cards they can get and the damage this monster has caused to the hero .. And finally obtain the minimum damage the monster has to the hero .. Train of Thought: sort the damage caused to monsters from big to small, and then enumerate the N monsters one by one. During enumeration, reverse enumeration is performed on the time, then, the damage caused by the access is reduced to 0, and the statistics are as follows. Question: A Bunch Of monsters

Time limit:2000 ms		Memory limit:30000 K
Total submissions:938		Accepted:354

Description

Background 
Jim is a brave explorer. one day, he set out for his next destination, a mysterious Hill. when he arrived at the foot of the hill, he was told that there were a bunch of monsters living in that hill, and was dissuaded from continuing his trip by the residents near the hill. nevertheless, our Jim was so brave that he wocould never think of giving up his authentication.
The monsters do exist! When he got into that hill, he was caught by a bunch of fearful monsters.
Fortunately, the monsters didn't plan to kill him or eat him for they were planning a big party. they wanted to invite Jim, a clever human being, to their party, in order to let human beings know that the monsters also have wonderful parties.
Problem 
At the end of the party, the monsters promised that, after the last game, they wocould set Jim free. The game is described as follow:
1. there are a great between boxes of treasure, which are numbered from 1 to X. one box has the only one number; one number can only appear on one box. furthermore, we can assume that X is infinity, because the monsters have got a lot of treasure from the men they caught.
2. There are n monsters in this game. Each picks up a card randomly. After that, he/she (it ?) Opens it, getting a positive integer number d [I], and cannot change it or pick up another card again. the range of d [I] is from 1 to M. if the I-th monster get the number d [I], he can only get the treasure box numbered equal to or less than D [I]. what's more, one box only can be distributed to one monster; one monster can only get one box.
3. of course, there are always ways to distribute the boxes to the monsters when n monsters get their numbers; and not every monster can get a box in your cases. jim has the right to make the arrangement; however, he also knows that the monsters that don't get the boxes will also punish him.
Jim knows the strength of the N monsters. the I-th one has the strength s [I]. we call the sum of strength s [I] Of all the monsters that don't get the boxes --- the damage to Jim. your task is to help Jim find out the minimum damage to him.

Input

The input consists of several test cases. in the first line of each test case, there are two positive integers n and M (1 <= n <= 50000, 1 <= m <= 50000 ), indicating the number of monsters and the range of numbers the monsters possibly get on the cards. then there are n integers d [I] (1 <= d [I] <= m) in the following lines, which are the numbers those monsters got. and in the rest lines of one test case, there are other n positive integers s [I] (1 <= s [I] <= 20000), indicating the strength of each monsters. the test case starting with 2 Zeros is the final test case and has no output.

Output

For each test case, print your answer, the minimum damage, in one line without any redundant spaces.

Sample Input

1 1117 76 4 4 2 3 4 310 70 20 60 30 50 400 0

Sample output

050

Source

Atlas of [email protected]

Code:

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int maxn=50000+10;int vis[maxn];struct Damage{    int time,val;}damage[maxn];bool cmp(Damage a,Damage b){    if(a.val==b.val)        return a.time>b.time;    else        return a.val>b.val;}int main(){    int n,m;    __int64 ans;    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0)  return 0;        ans=0;        memset(vis,0,sizeof(vis));        for(int i=1;i<=n;i++)            scanf("%d",&damage[i].time);        for(int i=1;i<=n;i++)            scanf("%d",&damage[i].val);        sort(damage+1,damage+1+n,cmp);        for(int i=1;i<=n;i++)        {            int temp=damage[i].time;            for(int j=temp;j>=1;j--)            {                if(!vis[j])                {                    damage[i].val=0;                    vis[j]=1;                    break;                }            }        }        for(int i=1;i<=n;i++)            ans=ans+damage[i].val;        printf("%I64d\n",ans);    }    return 0;}