bzoj 1122 [POI2008]账本BBB 模拟贪心，单调队列

[POI2008]账本BBB

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 524  Solved: 251
[Submit][Status][Discuss]

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

HINT

若p+序列和≠q，可以发现取反操作数量是确定的

尽量在前面做加法，后面做减法

旋转操作，暴力想法是把最后一位提前，就相当于连成一个环，在环上求值

所以在环上枚举起点

暴力走一遍环单纯是为了解决非负的问题

在环上用单调队列求一遍最小值，再枚举起点做无旋转的修改

 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<cstdio>
 #include<queue>

 #define N 1000007
 #define ll long long
 inline ll read()
 {
     ll x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 ll n,p,q,x,y;
 int que[N<<],seq[N<<],mn[N];
 char ch[N];

 inline ll max(ll a,ll b)
 {
     if (a>b) return a;
     else return b;
 }
 inline ll min(ll a,ll b)
 {
     if (a>b) return b;
     else return a;
 }
 inline ll abs(ll a)
 {
     if (a>) return a;
     else return -a;
 }
 int main()
 {
     n=read(),p=read(),q=read(),x=read(),y=read();
     scanf("%s",ch+);
     for (int i=n<<;i>n;i--)
         seq[i]=seq[i+]+(ch[i-n]=='+'?:-);
     for (int i=n;i>=;i--)
         seq[i]=seq[i+]+(ch[i]=='+'?:-);
     int hd=,tl=;
     for (int i=n<<;i>=;i--)
     {
         while(hd<=tl&&seq[i]>seq[que[tl]]) tl--;
         que[++tl]=i;
         while(hd<=tl&&seq[hd]-i>=n) hd++;
         if (i<=n) mn[i]=seq[i]-seq[que[hd]];
     }
     ll total=seq[n+],tmp=(q-p-total)/,ans=1e16,res;
     for (int i=;i<n;i++)
     {
         res=x*abs(tmp)+y*(ll)i;
         if (i==)
         {
             mn[]+=p+max(tmp,)*;
             if (mn[]<) res+=*x*((-mn[])/);
         }
         else
         {
             mn[n-i+]+=p+max(tmp,)*;
             if (mn[n-i+]<) res+=*x*((-mn[n-i+])/);
         }
         ans=min(ans,res);
     }
     printf("%lld\n",ans);
 }