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CodeForces 377B---Preparing for the Contest(二分+貪心)

來源:程序員人生   發布時間:2014-11-24 08:34:41 閱讀次數:2197次
C - Preparing for the Contest
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 377B

 

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.

題意給出m個bug,每一個bug有個復雜程度,有n個同學每一個同學有自己的能力值b,和想要的東西c,

如果雇傭第i個同學,那末能解決所有復雜程度小于等于b[i]的bug,每天1人只能解決1個,學校要付出c,不論i解決了幾個bug

問,學校在付出不超過s,且最少的天數需要多少。

有兩個限制,1.總和不能超過s,2.要求最少天數。

只能限制1個,來求另外一個,如果求總和不能超過s,不好求,那末只能求最少天數,2分枚舉最少的天數,找出最小花費,得到最后的結果。

如果是時間為t,那末找出所有能力大于當前最大的bug的人,找出需要c最少的,使用優先隊列保護,讓找出的人工作t天,工作bug最大的t個,使得后面的bug可以找更多的人來修。

 

 

#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; #define LL __int64 #define INF 0x3f3f3f3f struct node { LL b , c , i ; // bool operator < (const node &x) const { // return c > x.c ; // } friend bool operator< (node n1, node n2) { return n1.c > n2.c; } } p[200000] , q ; priority_queue <node> que ; struct node1 { LL i , a ; } bug[200000]; bool cmp(node x,node y) { return x.b > y.b ; } bool cmp1(node1 x,node1 y) { return x.a > y.a ; } LL last[110000] , now[110000 ] , n , m ,s ; LL f(LL t) { while( !que.empty() ) que.pop(); LL i = 0 , j = 0 , ans = 0 , k ; while(j < m) { while(i < n && p[i].b >= bug[j].a) { que.push( p[i] ) ; i++ ; } if( que.empty() ) return s+1 ; q = que.top(); que.pop(); ans += q.c ; k = j+t ; while(j < m && j < k) { now[ bug[j].i ] = q.i ; j++ ; } } return ans ; } int main() { LL i , j ; memset(last,⑴,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,cmp1); for(i = 0 ; i < n ; i++) { scanf("%I64d", &p[i].b); p[i].i = i+1 ; } for(i = 0 ; i < n ; i++) { scanf("%I64d", &p[i].c); } sort(p,p+n,cmp); LL 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] = now[i] ; high = mid⑴ ; } else low = mid+1 ; } if( last[0] == ⑴ ) printf("NO "); else { printf("YES "); for(i = 0 ; i < m ; i++) { if(i == m) printf("%d ", last[i]); else printf("%d ", last[i]); } } return 0; }


 

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