标签：main repr min log htm 后缀 limit length eve
| Time Limit: 2000MS | Memory Limit: 65536K | |
Description
Input
Output
Sample Input
3 baaaababababbababbab 11 baaaababababbababbab 3 cccccc 0
Sample Output
5 12 none 4 2
分析：求出现至少m次的最长字符子串及最大位置；
后缀数组+set；（二分hash）
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <unordered_map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, ls[rt] #define Rson mid+1, R, rs[rt] #define sys system("pause") #define freopen freopen("in.txt","r",stdin) const int maxn=4e4+10; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } int n,m,k,t,cntA[maxn],cntB[maxn],sa[maxn],lev[maxn],height[maxn],A[maxn],B[maxn],tsa[maxn]; char ch[maxn]; void solve() { for (int i = 0; i < 256; i ++) cntA[i] = 0; for (int i = 1; i <= n; i ++) cntA[ch[i]] ++; for (int i = 1; i < 256; i ++) cntA[i] += cntA[i - 1]; for (int i = n; i; i --) sa[cntA[ch[i]] --] = i; lev[sa[1]] = 1; for (int i = 2; i <= n; i ++) { lev[sa[i]] = lev[sa[i - 1]]; if (ch[sa[i]] != ch[sa[i - 1]]) lev[sa[i]] ++; } for (int l = 1; lev[sa[n]] < n; l <<= 1) { for (int i = 0; i <= n; i ++) cntA[i] = 0; for (int i = 0; i <= n; i ++) cntB[i] = 0; for (int i = 1; i <= n; i ++) { cntA[A[i] = lev[i]] ++; cntB[B[i] = (i + l <= n) ? lev[i + l] : 0] ++; } for (int i = 1; i <= n; i ++) cntB[i] += cntB[i - 1]; for (int i = n; i; i --) tsa[cntB[B[i]] --] = i; for (int i = 1; i <= n; i ++) cntA[i] += cntA[i - 1]; for (int i = n; i; i --) sa[cntA[A[tsa[i]]] --] = tsa[i]; lev[sa[1]] = 1; for (int i = 2; i <= n; i ++) { lev[sa[i]] = lev[sa[i - 1]]; if (A[sa[i]] != A[sa[i - 1]] || B[sa[i]] != B[sa[i - 1]]) lev[sa[i]] ++; } } for (int i = 1, j = 0; i <= n; i ++) { if (j) j --; while (ch[i + j] == ch[sa[lev[i] - 1] + j]) j ++; height[lev[i]] = j; } } multiset<int>p; set<int>q; int main() { int i,j; while(~scanf("%d",&m)&&m) { scanf("%s",ch+1); n=strlen(ch+1); if(m==1) { printf("%d %d\n",n,0); continue; } solve(); p.clear(),q.clear(); rep(i,1,m-1)p.insert(height[i]),q.insert(sa[i]); int ans=0,pos=-1; rep(i,m,n) { p.erase(p.lower_bound(height[i-m+1])); p.insert(height[i]); q.insert(sa[i]); if(ans<=*p.begin()) { auto x=q.end(); x--; if(ans<*p.begin()||*x-1>pos)pos=*x-1; ans=*p.begin(); } q.erase(sa[i-m+1]); } if(ans)printf("%d %d\n",ans,pos); else puts("none"); } //system("Pause"); return 0; }
标签：main repr min log htm 后缀 limit length eve
原文地址：http://www.cnblogs.com/dyzll/p/6010203.html
