HDU6261：Rikka with Mutex（二分 + 貪心）

Rikka with Mutex

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 132    Accepted Submission(s): 28

 

Problem Description

Sometimes, technical terms implicate some life philosophy. Mutex is one of them. On your way to dream, you may be locked by some difficulties, and you need someone to stop his step, and help you get through them.

To help you know better about the life philosophy inside mutex, Rikka comes up with a simple task. Maybe some of you know little about mutex, so she uses another scene to replace it.

There are n gates in a row, several people in the left side of the gates and all of them want to go to the right side. There are two kinds of gates: black and white. These people share energy, which is represented by a non-negative number E. Initially, E=0.

If one person walks through a white gate, he will gain one point of energy, i.e., E will be added by 1. And if one person walks through a black gate, he will lose one point of energy, i.e., E will be subtracted by 1. Since E must be a non-negative integer, if E=0, no one can walk through a black gate until someone walks through a white gate. You can assume there won't be two people moving at the same time and all the people are selfless.

We use P to represent a black gate, V to represent a white gate and use a PV string to represent the row. Initially, all the people are at the beginning of the string, and all of them want to go through the whole string. But unfortunately, sometimes it may be impossible. So, they want to send at least one person to the right side. 

Your task is to find out the minimal number of people which this group needs to achieve this goal.

For example, if the row is VPP, they need at least two people: The first person walk through the first white gate and the second person can use this point of energy to go through the whole string.

 

 

Input

The first line contains a single numner t(1≤t≤103), the number of the testcases.

For each testcase, the first line contains a PV string s(1≤|s|≤105) describing the gates.

The input guarantees that there are at most 30 testcases with |S|>1000.

 

 

Output

For each testcase, output a single integer, the answer. And if it is impossible, output −1.
 

 

 

Sample Input

4 VPP VPPVVVVPPPPPPPP VPPPPPPPPPPPPPP P

 

 

Sample Output

2 3 14 -1

題意：給一個字符串，一個人從第一個出發到最後一個，其中遇到V生命值加上1，遇到P生命值減去1，任何時候生命都不允許是負數，問初始至少要安排幾個人，才能使得至少有一個人能夠到達終點（所有人共用生命值）。

思路：冬令營的題目，首先考慮二分答案，怎麼選擇驗證的策略呢？假如字符串是VVVVVPPPVVPP，一開始肯定是所有人走到第五個V處獲得最多的生命值，記爲tot，然後此時判斷生命值足不足夠派一個人去到終點，如果還不足夠，就派一個人自己往前走直到生命值比tot大就停止，然後剩下的人全部跟上來。此策略可以保證生命值最多，因爲如果派一個人往前走tot不斷減少，再派其他人跟上來也許會使情況更糟糕，如果一直沒遇到比tot大的點，說明走不到終點。

這題當時沒做出來，要反思一下二分答案的精髓，其實二分一個答案mid，check的時候只要考慮 如果答案是mid，這mid個全部用完它，按照某個策略一定能完成或者如果不是這樣就一定完不成。而不是check的時候還去考慮“標準”的完成方法。

# include <iostream>
# include <cstring>
# include <cstdio>
using namespace std;
char s[100003];
int len;
bool ok(int x)
{
    int tot = 0, pre = 0, one = 0;
    while(one < len)
    {
        ++one;
        tot += (s[one]=='P'?-1:1);
        if(tot < 0) return false;
        else if(tot > pre) tot = tot+(tot-pre)*(x-1), pre=tot;
    }
    return true;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s+1);
        len = strlen(s+1);
        if(s[1] == 'P') puts("-1");
        else
        {
            int l=1, r=len;
            while(l <= r)
            {
                int mid = l + r >> 1;
                if(ok(mid)) r = mid - 1;
                else l = mid + 1;
            }
            printf("%d\n",l);
        }
    }
    return 0;
}