BZOJ 2565 Mancher 算法求 最长双回文串

http://www.lydsy.com/JudgeOnline/problem.php?id=2565

求出最长的子串可以分成两个回文串

先处理出以每个位置为中心的最长回文子串，然后我是求了两个这样的数组left[] right[]，分别表示某位置往左最长的回文子串（以该位置结尾），右边的雷同

需要线性扫两遍。。。

处理这两个数组的时候细节问题整了半天，所幸最后还是整出来了。

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int maxn= 200010;namespace M {int n;struct node {int a,b;int cen;node() {}node(int _a,int _b):a(_a),b(_b){};bool operator < (const node&cmp) const {return b < cmp.b;}}in[maxn];vector<pair<int,int> > P[maxn] ;///P[i]记录以i为中心的所有的最长回文子串（其实最多只有两个,奇偶性。。） int left[maxn] , right[maxn];bool ji[maxn];void solve(){memset(ji,false,sizeof(ji));int mx = 0;for(int i = 0; i < n; i++)mx = max(mx,in[i].b);for(int i = 1; i <= mx; i++) P[i].clear();for(int i = 0; i < n; i++){in[i].cen= in[i].a+in[i].b >> 1;P[in[i].cen].push_back(make_pair(in[i].a,in[i].b));}        int pt = 1;        for(int i = 1; i <= mx; i++)        {        for(; pt < i; pt++)        {        int a = P[pt][0].first;        int b = P[pt][0].second;        if(b>=i)        {        ji[i] = true;        break;}if(P[pt].size()==1) continue;a = P[pt][1].first;b = P[pt][1].second;if(b>=i){break;}}left[i] = pt;}for(int i = 1; i <= mx; i++){left[i] = max(1,(i - left[i])*2+ji[i]);}pt = mx;memset(ji,false,sizeof(ji));for(int i = mx; i >= 1; i--){right[i] = i;         for(; pt >= i; pt--)        {            int a, b;    if(P[pt].size() > 1)    {                a = P[pt][1].first;   b = P[pt][1].second;   if(a<=i)    {                      right[i] = pt + 1;  break;   }    }        a = P[pt][0].first;        b = P[pt][0].second;        if(a<=i)        {right[i] = pt;        ji[i] = true;        break;}}}for(int i = 1; i <= mx; i++){right[i] = max(1,(right[i] - i)*2+ji[i]);} int ans = 0;for(int i = 1; i < mx; i++){               ans = max(ans,left[i]+right[i+1]);}printf("%d\n",ans);}}struct Mancher {char str[maxn];//start from index 1int p[maxn];char s[maxn];int n;void checkmax(int &ans,int b){if(b>ans) ans=b;}inline int min(int a,int b){return a<b?a:b;}void kp(){int i;int mx = 0;int id;for(i=1; i<n; i++){if( mx > i )p[i] = min( p[2*id-i], p[id]+id-i );elsep[i] = 1;for(; str[i+p[i]] == str[i-p[i]]; p[i]++) ;if( p[i] + i > mx ) {mx = p[i] + i;id = i;}}}void pre(){int i,j,k;n = strlen(s);str[0] = '$';str[1] = '#';for(i=0;i<n;i++){str[i*2 + 2] = s[i];str[i*2 + 3] = '#';}n = n*2 + 2;str[n] = 0;}void solve() // 求出所有的最长回文子串所在的区间{int & tot = M::n;tot = 0;for(int i = 2; i < n; i++){if(i%2&&p[i]==1) continue;if(i%2){M::in[tot++] = M::node(i/2-p[i]/2+1,i/2+p[i]/2);//printf("%d %d\n",i/2-p[i]/2+1,i/2+p[i]/2);}else {M::in[tot++] = M::node(i/2-(p[i]/2-1),i/2+(p[i]/2-1));//printf("%d %d\n",i/2-(p[i]/2-1),i/2+(p[i]/2-1));}}}}task1;int main(){while( scanf("%s", task1.s) !=EOF ){task1.pre();task1.kp();task1.solve();M::solve();}return 0;}