bzoj1122: [POI2008]账本BBB

Description

一个长度为n的记账单，+表示存￥1，-表示取￥1。现在发现记账单有问题。一开始本来已经存了￥p，并且知道最后账户上还有￥q。你要把记账单修改正确，使得 1：账户永远不会出现负数； 2：最后账户上还有￥q。你有2种操作： 1：对某一位取反，耗时x； 2：把最后一位移到第一位，耗时y。

Input

The first line contains 5 integers n, p, q, x and y (1  n  1000000, 0  p;q  1000000, 1  x;y  1000), separated by single spaces and denoting respectively: the number of transactions done by Byteasar, initial and final account balance and the number of seconds needed to perform a single turn (change of sign) and move of transaction to the beginning. The second line contains a sequence of n signs (each a plus or a minus), with no spaces in-between. 1 ≤ n ≤ 1000000, 0 ≤ p ,q ≤ 1000000, 1 ≤x,y ≤ 1000)

Output

修改消耗的时间

Sample Input

9 2 3 2 1
---++++++

Sample Output

3

 

题解：

http://blog.csdn.net/wzq_qwq/article/details/48528159

code：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
char ch;
bool ok;
void read(int &x){
    for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    if (ok) x=-x;
}
const int maxn=2000005;
typedef long long int64;
char s[maxn];
int n,p,q,x,y,sum[maxn],minv[maxn],tmp,head,tail;
int64 ans=1LL<<60;
struct Data{
    int val,id;
}que[maxn];
int main(){
    read(n),read(p),read(q),read(x),read(y);
    scanf("%s",s+1);
    for (int i=n<<1;i>n;i--) sum[i]=sum[i+1]+(s[i-n]=='+'?1:-1);
    for (int i=n;i;i--) sum[i]=sum[i+1]+(s[i]=='+'?1:-1);
    head=1,tail=0;
    for (int i=n<<1;i;i--){
        while (head<=tail&&que[head].id>i+n) head++;
        while (head<=tail&&que[tail].val<sum[i]) tail--;
        que[++tail]=(Data){sum[i],i};
        minv[i]=sum[i]-que[head].val;
    }
    tmp=(q-p-sum[n+1])>>1;
    for (int i=1;i<=n;i++){
        int64 res=1LL*(n-i+1)%n*y+abs(tmp)*x;
        minv[i]+=p+max(tmp,0)*2;
        if (minv[i]<0) res+=((1-minv[i])>>1)*x*2;
        ans=min(ans,res);
    }
    printf("%lld
",ans);
    return 0;
}