51nod 1391 01 String (Exercise thinking good question)

Topic http://www.51nod.com/onlineJudge/questionCode.html#problemId=1391?iceId=20917

1391 01 Strings Source: codility Base time limit: 1 seconds space limit: 131072 KB Score: 40 Difficulty: 4-level algorithm problem

Given a 01 string s, find a substring of it as long as possible s[I.. J], satisfies the existence of a position i<=x <=j, s[i. X] in more than 0:1, and S[x + 1..J] in more than 1:0. The length of the oldest string that satisfies the condition. Input

A row contains a string s that is composed only of 0 and 1. The length of S is not more than 1000000.

Output

A row contains an integer that represents the length of the oldest string that satisfies the requirement.

Input example

10

Output example

0

Problem Solving Ideas:

Very interesting a topic, see the topic easy to let us think, 2 times pre-treatment, 1 times in the past to find out to this position 0 of the number is greater than the maximum of 1

Length, and one time from the trip to find the number of this position 1 is greater than 0. But how to find it? Then we need a little skill, we let 0 generations

Table -1,1 represents 1, then a paragraph and less than 0, then 0 of the number is greater than 1, greater than 0 represents the number of 1 is greater than the number of 0, so it is converted into this position

The maximum length is greater than 0 or less than 0.

so far, we can not solve the problem, we take from the post-processing as an example, when from the starting position to here and less than 0 when, well disposed of, this long

The degree is, but when and greater than equals 0 o'clock what? Assuming that the value here is cur (cur>=0), see if the front appears cur+1, cur+1, the earliest

Now cur+1 position to the current value is to meet the number of 0 is greater than 1 of the number of the longest length, why the cur+2,cur+3 also meet the requirements, but from

it The beginning of some values must be less than the length from cur+1, certain, because cur+2,cur+3 must be in the back of the cur+1, so the topic can be done

The

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace     std;const int maxn=1000000+1000;int h[maxn];int a[maxn];int b[maxn];int c[maxn];char s[maxn];int Main () {scanf ("%s", s);    int N=strlen (s);    int cur=0;        for (int i=0;i<n;i++) {if (s[i]== ' 0 ') a[i]=-1;    else a[i]=1;    } memset (H,-1,sizeof (h));    memset (B,0,sizeof (a));        for (int i=0;i<n;i++) {cur+=a[i];        if (cur<0) b[i]=i+1;            else {if (h[cur+1]!=-1) {b[i]=i-h[cur+1];                } else {h[cur]=i;            b[i]=0;    }}} memset (H,-1,sizeof (h));    Memset (C,0,sizeof (c));    cur=0;        for (int i=n-1;i>=0;i--) {cur+=a[i];        if (cur>0) c[i]=n-i;            else {if (h[-(cur-1)]!=-1) {c[i]=h[-(cur-1)]-i;            }else {c[i]=0;            h[-(cur)]=i;    }}} int ans=0;    for (int i=0;i<n;i++) {if (b[i]>0&&c[i+1]>0) Ans=max (B[i]+c[i+1],ans);    } cout<<ans<<endl; return 0;}

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51nod 1391 01 String (Exercise thinking good question)