记得KPM出的题中有过类似的。。

先求出前缀,然后符合题意的条件的i和j满足a[i]-a[j]==b[i]-b[j]==c[i]-c[j]

等价转换得a[i]-b[i]==a[j]-b[j]&&a[i]-c[i]==a[j]-c[j]

所以a[i]-b[i]和a[i]-c[i]用hash记录就可以了(我懒直接用map了

 1 #include<bits/stdc++.h>
 2 #define inc(i,l,r) for(int i=l;i<=r;i++)
 3 #define dec(i,l,r) for(int i=l;i>=r;i--)
 4 #define link(x) for(edge *j=h[x];j;j=j->next)
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define inf 1e9
 7 #define ll long long
 8 #define succ(x) (1<<x)
 9 #define lowbit(x) (x&(-x))
10 #define NM 200000+5
11 using namespace std;
12 int read(){
13     int x=0,f=1;char ch=getchar();
14     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
15     while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
16     return x*f;
17 }
18 char st[NM];
19 int n,a[NM],b[NM],c[NM],ans,tot;
20 map<pair<int,int>,int>h;
21 int main(){
22     freopen("data.in","r",stdin);
23     n=read();
24     scanf("%s",st);
25     inc(i,1,n){
26         a[i]=a[i-1];b[i]=b[i-1];c[i]=c[i-1];
27         switch(st[i-1]){
28             case 'J':a[i]++;break;
29             case 'I':b[i]++;break;
30             case 'O':c[i]++;break;
31         }
32         pair<int,int>p=make_pair(a[i]-b[i],a[i]-c[i]);
33         if(h[p])ans=max(ans,i-h[p]);else h[p]=i;
34         if(a[i]==b[i]&&b[i]==c[i])ans=max(ans,i);
35     }
36     printf("%d\n",ans);
37     return 0;
38 }
View Code

 

posted on 2016-03-08 21:36  onlyRP  阅读(122)  评论(0编辑  收藏  举报