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CodeForces 377B Preparing for the Contest 贪心（2分加优先队列）

分类: IT文章 • 2024-06-29 13:14:55
CodeForces 377B Preparing for the Contest 贪心（二分加优先队列）
题意给出m个bug，每一个bug有个复杂程度，有n个同学每一个同学有自己的能力值b，和想要的东西c，

假设雇佣第i个同学，那么能解决全部复杂程度小于等于b[i]的bug，每天一人仅仅能解决一个，学校要付出c，不论i攻克了几个bug

问，学校在付出不超过s，且最少的天数须要多少。

有两个限制，1.总和不能超过s，2.要求最少天数。

仅仅能限制一个，来求还有一个，假设求总和不能超过s，不好求，那么仅仅能求最少天数，二分枚举最少的天数，找出最小花费，得到最后的结果。

假设是时间为t，那么找出全部能力大于当前最大的bug的人，找出须要c最少的，使用优先队列维护，让找出的人工作t天，工作bug最大的t个，使得后面的bug能够找很多其它的人来修。

Description

Soon there will be held the world's largest programming contest, but the testing system still has m bugs. The contest organizer, a well-known university, has no choice but to attract university students to fix all the bugs. The university has n students able to perform such work. The students realize that they are the only hope of the organizers, so they don't want to work for free: the i-th student wants to getci 'passes' in his subjects (regardless of the volume of his work).

Bugs, like students, are not the same: every bug is characterized by complexity aj, and every student has the level of his abilities bi. Student i can fix a bug j only if the level of his abilities is not less than the complexity of the bug: bi ≥ aj, and he does it in one day. Otherwise, the bug will have to be fixed by another student. Of course, no student can work on a few bugs in one day. All bugs are not dependent on each other, so they can be corrected in any order, and different students can work simultaneously.

The university wants to fix all the bugs as quickly as possible, but giving the students the total of not more than s passes. Determine which students to use for that and come up with the schedule of work saying which student should fix which bug.

Input

The first line contains three space-separated integers: n, m and s (1 ≤ n, m ≤ 105, 0 ≤ s ≤ 109) — the number of students, the number of bugs in the system and the maximum number of passes the university is ready to give the students.

The next line contains m space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — the bugs' complexities.

The next line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ 109) — the levels of the students' abilities.

The next line contains n space-separated integers c1, c2, ..., cn (0 ≤ ci ≤ 109) — the numbers of the passes the students want to get for their help.

Output

If the university can't correct all bugs print "NO".

Otherwise, on the first line print "YES", and on the next line print m space-separated integers: the i-th of these numbers should equal the number of the student who corrects the i-th bug in the optimal answer. The bugs should be corrected as quickly as possible (you must spend the minimum number of days), and the total given passes mustn't exceed s. If there are multiple optimal answers, you can output any of them.

Sample Input

Input
3 4 9
1 3 1 2
2 1 3
4 3 6
Output
YES
2 3 2 3
Input
3 4 10
2 3 1 2
2 1 3
4 3 6
Output
YES
1 3 1 3
Input
3 4 9
2 3 1 2
2 1 3
4 3 6
Output
YES
3 3 2 3
Input
3 4 5
1 3 1 2
2 1 3
5 3 6
Output
NO

Hint

Consider the first sample.

The third student (with level 3) must fix the 2nd and 4th bugs (complexities 3 and 2 correspondingly) and the second student (with level 1) must fix the 1st and 3rd bugs (their complexity also equals 1). Fixing each bug takes one day for each student, so it takes 2 days to fix all bugs (the students can work in parallel).

The second student wants 3 passes for his assistance, the third student wants 6 passes. It meets the university's capabilities as it is ready to give at most 9 passes.


#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<ctype.h>
#include<queue>
using namespace std;
__int64 n,m,s;
struct node
{
    __int64 a,b,i;
    friend bool operator <( node n1, node n2)
    {
        return n1.b>n2.b;
    }
} p[200005],top;
struct w
{
    __int64 i,a;
} bug[200005];
int cmp(w x,w y)
{
    return x.a>y.a;
}
int cmp1(node x,node y)
{
    return x.a>y.a;
}
__int64 PP[110000],last[110000];
int F(__int64 t)
{
    priority_queue<node>q;
    while( !q.empty() ) q.pop();
    __int64 ans=0,i=0,j=0;
    while(j<m)
    {
        while(p[i].a>=bug[j].a&&i<n)
        {
            q.push(p[i]);
            i++;
        }
        if(q.empty())
            return s+1;
        top=q.top();
        q.pop();
        ans+=top.b;
        if(ans>s)
            return s+1;
        __int64 k=j+t;
        while(j<m&&j<k)
        {
            PP[bug[j].i]=top.i;
            j++;
        }
    }
    return ans;
}
int main()
{
    __int64 i , j ;
    memset(last,-1,sizeof(last));
    scanf("%I64d %I64d %I64d", &n, &m, &s);
    for(i = 0 ; i < m ; i++)
    {
        scanf("%I64d", &bug[i].a);
        bug[i].i = i ;
    }
    sort(bug,bug+m,cmp);
    for(i = 0 ; i < n ; i++)
    {
        scanf("%I64d", &p[i].a);
        p[i].i = i+1 ;
    }
    for(i = 0 ; i < n ; i++)
    {
        scanf("%I64d", &p[i].b);
    }
    sort(p,p+n,cmp1);
    __int64 low = 1 , mid , high = m , min1 ;
    while( low <= high )
    {
        mid = (low+high)/2 ;
        min1 = F(mid);
        if( min1 <= s )
        {
            for(i = 0 ; i < m ; i++)
                last[i] = PP[i] ;
            high = mid-1 ;
        }
        else
            low = mid+1 ;
    }
    if( last[m-1] == -1 )
        printf("NO\n");
    else
    {
        printf("YES\n");
        for(i = 0 ; i < m ; i++)
        {
            if(i == m)
                printf("%d\n", last[i]);
            else
                printf("%d ", last[i]);
        }
    }
    return 0;
}


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