Given a 0-1 string, find a substring as long as possible, which contains 0 equal to the number of 1.

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1393

The method is fascinating. Also looked at other people's thinking to come out.

First consider turning 0 all into-1. Then a prefix is counted and expressed in sum[i].

Then the substring starting from the starting point is valid as long as the value of Sum[i] equals 0.

If the substring starting point is not 0, then as long as two occurrences of sum[i], the two values must be a valid substring.

For example:

Label: 1 2 3 4 5

Value 0 0 1 1 0

Sum-1-2-1 0-1

I==4 time sum[i]==0, so 1-4 is a valid substring, length 4

sum[1]==sum[3]==sum[5]==-1, so (1,3], (1,5], (3,5) is a valid substring, the maximum value is also 4.

I use a map, in fact, not so troublesome, as long as a variable can replace the sum array, and another marker array can be used instead of map mapping.

1#include <iostream>2#include <string.h>3#include <stdio.h>4#include <map>5 using namespacestd;6 Chars[1000010];7 intt[1000010],sum[1000010];8 intMain ()9 {Ten     //freopen ("A.txt", "R", stdin); Onescanf"%s", s+1); A     intL=strlen (s+1); -      for(intI=1; i<=l;i++) -         if(s[i]=='0') t[i]=-1; the         Elset[i]=1; -     //for (int i=0;i<l;i++) printf ("%d", T[i]); -sum[1]=t[1]; -map<int,int>m; +m[sum[1]]=1; -     intmaxn=0; +      for(intI=2; i<=l;i++) A     { atsum[i]=sum[i-1]+T[i]; -         //printf ("%d\n", Sum[i]); -         if(m[sum[i]]!=0) Maxn=max (maxn,i-M[sum[i]]); -         Elsem[sum[i]]=i; -         if(sum[i]==0) maxn=Max (maxn,i); -     } inprintf"%d\n", MAXN); -     return 0; to}

Given a 0-1 string, find a substring as long as possible, which contains 0 equal to the number of 1.