CodeForces 30E Tricky and Clever Password（hash+manacher）

题意：把一个回文串拆成prefix, middle, suffix三部分，中间那部分必须得是奇数，prefix与suffix对称并且长度可为0，把这3部分放进一个串中，成为A+prefix+B+middle+C+suffix，要使回文串最长，输出分界的位置。

做法：首先求出以小于等于i为结尾的prefix能匹配的最长suffix是多长L[i]，可以看出当左边界(即以i为结尾的prefix)匹配成功，就移动右边界（suffix），符合单调性。一直移动到两者相交停止，所以可以利用hash，o(n)的求出这个（或者利用扩展kmp）。再利用manacher求出每个点为中心的最长回文串，枚举i，最大值就是max(2*w[i]-1+2*min(L[i-w[i]], n - (i+w[i])-1) )。w[i]代表以i为中心包括i到最右端回文串的长度。

总复杂度o(n)。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<queue>#include<set>#include<map>#include<cmath>#include<ctime>#include<string.h>#include<string>#include<sstream>#include<bitset>using namespace std;#define ll __int64#define ull unsigned long long#define eps 1e-8#define NMAX 1000000000#define MOD 51123987#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define PI acos(-1)#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x.end())template<class T>inline void scan_d(T &ret){    char c;    int flag = 0;    ret=0;    while(((c=getchar())<'0'||c>'9')&&c!='-');    if(c == '-')    {        flag = 1;        c = getchar();    }    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();    if(flag) ret = -ret;}template<class T> inline T Max(T a, T b){ return a > b ? a : b; }template<class T> inline T Min(T a, T b){ return a < b ? a : b; }const int X = 141;const int maxn = 200000+10;ull h[maxn],x[maxn];char s[maxn],s2[maxn];int n;void init(){    h[2*n+1] = 0;    for(int i = 2*n; i >= 1; i--) h[i] = h[i+1]*X+(ull)(s[i]-'a');    x[0] = 1;    for(int i = 1; i <= 2*n; i++) x[i] = x[i-1]*(ull)X;}int g[maxn],w[maxn];int fag[maxn];inline bool judge(int l, int r, int len){    ull t1 = h[l]-h[l+len]*x[len];    ull t2 = h[r]-h[r+len]*x[len];//    cout<<t1<<" "<<t2<<endl;    if(t1 == t2) return true;    return false;}int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o.txt","w",stdout);#endif    scanf("%s",s+1);    n = strlen(s+1);    for(int i = n+1; i <= 2*n; i++)        s[i] = s[2*n-i+1];//    s[2*n+1] = '\0';//    cout<<s+1<<endl;    init();    memset(g,0,sizeof(g));    int l = 2*n,r = n;    while(l > 2*n-r+1)    {        if(judge(r,l,n-r+1))        {            g[2*n-l+1] = n-r+1;            r--;            l--;        }        else l--;    }    fag[1] = 1;    for(int i = 2; i <= n; i++)    {        if(g[i] > g[i-1])        {            fag[i] = i;        }        else        {            g[i] = g[i-1];            fag[i] = fag[i-1];        }    }    s2[0] = '$'; s2[n+1] = '#';    for(int i = 1; i <= n; i++)        s2[i] = s[i];    int p0,p = 0;    for(int i = 1; i <= n; i++)    {        if(p > i)        {            if(w[2*p0-i] < p-i+1) w[i] = w[2*p0-i];            else            {                int j;                for(j = p-i; s2[i+j] == s2[i-j]; j++);                w[i] = j;                p0 = i;                p = i+j-1;            }        }        else        {            int j;            for(j = 1; s2[i+j] == s2[i-j]; j++);            w[i] = j;            p0 = i;            p = i+j-1;        }    }    int ans = 0,re[3][2];    for(int i = 1; i <= n; i++)    {        if(ans < 2*w[i]-1+2*Min(g[i-w[i]],n-(i+w[i])+1))        {            re[1][0] = i-w[i]+1;            re[1][1] = 2*w[i]-1;            if(g[i-w[i]] > n-(i+w[i])+1)            {                re[0][0] = fag[i-w[i]]-n+i+w[i];                re[0][1] = re[2][1] = n-(i+w[i])+1;                re[2][0] = i+w[i];            }            else            {                re[0][0] = fag[i-w[i]]-g[i-w[i]]+1;                re[0][1] = re[2][1] = g[i-w[i]];                re[2][0] = n-g[i-w[i]]+1;            }            ans = 2*w[i]-1+2*Min(g[i-w[i]],n-(i+w[i])+1);        }    }    int nct = 3;    if(re[0][0]+re[0][1] == re[1][0] && re[0][0]+re[0][1]+re[1][1] == re[2][0])    {        re[0][1] += re[1][1]+re[2][1];        nct = 1;    }    if(re[0][1] == 0 && re[2][1] == 0)    {        re[0][0] = re[1][0];        re[0][1] = re[1][1];        nct = 1;    }    printf("%d\n",nct);    for(int i = 0; i < nct; i++)        printf("%d %d\n",re[i][0],re[i][1]);//    printf("%d\n",ans);    return 0;}