18: Array C
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 586 Solved: 104
[Submit][Status][Web Board]
Description
Giving two integers and and two arrays and both with length , you should construct an array also with length which satisfied:
1.0≤Ci≤Ai(1≤i≤n)
2..png){kind=small}
and make the value S be minimum. The value S is defined as:
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Input
There are multiple test cases. In each test case, the first line contains two integers n(1≤n≤1000) andm(1≤m≤100000). Then two lines followed, each line contains n integers separated by spaces, indicating the array Aand B in order. You can assume that 1≤Ai≤100 and 1≤Bi≤10000 for each i from 1 to n, and there must be at least one solution for array C. The input will end by EOF.
Output
For each test case, output the minimum value S as the answer in one line.
Sample Input
3 4
2 3 4
1 1 1
Sample Output
6
HINT
In the sample, you can construct an array [1,1,2](of course [1,2,1] and [2,1,1] are also correct), and is 6.
题目大意：给你n，m。表示a数组和b数组。然后让你选m个数组成c数组，保证c[i] <= a[i]，要求sigma(c[i]*c[i]*b[i])求和最小。可能没描述不太清楚，自己再看看题吧。。。
解题思路：自己想了一个mlogn的做法，但是一直超时到死。。。赛后听大牛的解法，的确很巧妙ORZ。
自己的超时想法：
#include<stdio.h>
#include<algorithm>
//#include<string.h>
//#include<math.h>
//#include<string>
//#include<iostream>
#include<queue>
//#include<stack>
//#include<map>
//#include<vector>
//#include<set>
using namespace std;
typedef long long LL;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn = 1e3 + 30;
const LL INF = 0x3f3f3f3f;
const LL mod = 9973;
typedef long long LL;
typedef unsigned long long ULL;
struct Num{
int b, c, prod, id;
bool operator < (const Num & rhs)const{
return prod > rhs.prod;
}
}nums[maxn];
int A[maxn], B[maxn], ans[maxn];
priority_queue<Num>PQ;
int Scan() { //输入外挂
int res = 0, flag = 0;
char ch;
if((ch = getchar()) == '-') flag = 1;
else if(ch >= '0' && ch <= '9') res = ch - '0';
while((ch = getchar()) >= '0' && ch <= '9')
res = res * 10 + (ch - '0');
return flag ? -res : res;
}
void Out(int a) { //输出外挂
if(a < 0) { putchar('-'); a = -a; }
if(a >= 10) Out(a / 10);
putchar(a % 10 + '0');
}
int main(){
int n,m;
while(scanf("%d %d",&n,&m)!=EOF){
while(!PQ.empty()) PQ.pop();
for(int i = 1; i <= n; ++i){
scanf("%d",&A[i]);
// A[i] = Scan();
nums[i].c = 1;
}
for(int i = 1; i <= n; i++){
scanf("%d",&nums[i].b);
// nums[i].b = Scan();
nums[i].id = i;
nums[i].prod = nums[i].b * (nums[i].c*nums[i].c);
PQ.push(nums[i]);
}
Num nm = PQ.top(); PQ.pop();
ans[nm.id]++;
Num tmp;
for(int i = 1; i < m; ++i){
if(PQ.empty()){
ans[nm.id] += m-i; break;
}
// cout<<PQ.size()<<" ++++"<<endl;
tmp = PQ.top();
// printf("%d %d %d++
",tmp.id,tmp.c,tmp.prod);
nm.c++;
nm.prod = nm.b*nm.c*nm.c;
if(nm.c > A[nm.id]){
PQ.pop();
nm = tmp;
ans[nm.id] = nm.c; continue;
}
if(tmp.prod < nm.prod){
PQ.pop();
PQ.push(nm);
nm = tmp;
ans[nm.id] = nm.c;
}else{
ans[nm.id] = nm.c;
}
}
int res = 0;
for(int i = 1; i <= n; ++i){
// printf("%d
",ans[i]);
res += nums[i].b * ans[i] * ans[i];
ans[i] = 0;
}
//printf("%d
",res);
Out(res);
puts("");
}
return 0;
}
大牛的解法：
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
using namespace std;
typedef long long LL;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn = 1e6 + 30;
const LL INF = 0x3f3f3f3f;
const LL mod = 9973;
typedef long long LL;
typedef unsigned long long ULL;
int a[maxn],b[maxn] ,cost[maxn];
int main(){
int n, m;
while(scanf("%d%d",&n,&m)!=EOF){
for(int i = 1; i <= n; ++i){
scanf("%d",&a[i]);
}
for(int i = 1; i <= n; ++i){
scanf("%d",&b[i]);
}
int cnt = 0;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= a[i]; ++j){
cost[cnt] = (2*j-1)*b[i];
cnt++;
}
}
sort(cost,cost+cnt);
LL res = 0;
for(int i = 0; i < m; ++i){
res += cost[i];
}
printf("%lld
",res);
}
return 0;
}