Common Substring poj3415

 弱的无奈，看论文后想不到维护的方法，看题解知道了维护的方法自己写又WA的一B，最后照搬了了别人的代码==。。。维护方法见代码和注释吧

这题除了用栈来维护，还有一种用笛卡尔树(就是一种特殊的堆，黑书P94)来统计的方法也是，按照height的值建树(O(n))后统计一遍(O(N))，统计的方法是，对于height >= K的节点，

∑(左子树A后缀的个数×右子树B后缀的个数+左子树B后缀的个数×右子树A后缀的个数+与该节点不属于同一串的子节点的个数)×(height-K+1)

不过统计时可能会爆栈

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <cstring>#include <stack>#include <cctype>#include <utility>   #include <map>#include <string>  #include <climits> #include <set>#include <string>    #include <sstream>#include <utility>   #include <ctime>#include <bitset>using std::priority_queue;using std::vector;using std::swap;using std::stack;using std::sort;using std::max;using std::min;using std::pair;using std::map;using std::string;using std::cin;using std::cout;using std::set;using std::queue;using std::string;using std::stringstream;using std::make_pair;using std::getline;using std::greater;using std::endl;using std::multimap;using std::deque;using std::unique;using std::lower_bound;using std::random_shuffle;using std::bitset;typedef long long LL;typedef unsigned long long ULL;typedef pair<int, int> PAIR;typedef multimap<int, int> MMAP;typedef LL TY;const int MAXN(200010);const int MAXM(50010);const int MAXE(100010);const int MAXK(6);const int HSIZE(131313);const int SIGMA_SIZE(52);const int MAXH(19);const int INFI((INT_MAX-1) >> 1);const ULL BASE(31);const LL LIM(10000000);const int INV(-10000);const int MOD(1000000007);const double EPS(1e-7);template<typename T> void checkmax(T &a, T b){if(b > a) a = b;}template<typename T> void checkmin(T &a, T b){if(b < a) a = b;}template<typename T> T ABS(const T &a){return a < 0? -a: a;}int K;struct SA{char S[MAXN];int sa[MAXN], t1[MAXN], t2[MAXN], cnt[MAXN], len, M;void init(int tl, int tm = 128){len = tl;M = tm;int *p1 = t1;int *p2 = t2;for(int i = 0; i < M; ++i) cnt[i] = 0;for(int i = 0; i <= len; ++i) ++cnt[p1[i] = S[i]];for(int i = 1; i < M; ++i) cnt[i] += cnt[i-1];for(int i = len; i >= 0; --i) sa[--cnt[p1[i]]] = i;int temp = 1;for(int k = 1; temp <= len; k <<= 1){temp = 0;for(int i = len-k+1; i <= len; ++i) p2[temp++] = i;for(int i = 0; i <= len; ++i)if(sa[i] >= k)p2[temp++] = sa[i]-k;for(int i = 0; i < M; ++i) cnt[i] = 0;for(int i = 0; i <= len; ++i) ++cnt[p1[p2[i]]];for(int i = 1; i < M; ++i) cnt[i] += cnt[i-1];for(int i = len; i >= 0; --i) sa[--cnt[p1[p2[i]]]] = p2[i];swap(p1, p2);temp = 1;p1[sa[0]] = 0;for(int i = 1; i <= len; ++i)p1[sa[i]] = p2[sa[i-1]] == p2[sa[i]] && p2[sa[i-1]+k] == p2[sa[i]+k]? temp-1: temp++;M = temp;}}int rank[MAXN], height[MAXN];void getHeight(){int k = 0;for(int i = 0; i <= len; ++i)rank[sa[i]] = i;for(int i = 0; i < len; ++i){if(k) --k;int j = sa[rank[i]-1];while(S[i+k] == S[j+k]) ++k;height[rank[i]] = k;}}/*int Log[MAXN];int table[MAXH][MAXN];void initLog(){Log[0] = -1;for(int i = 1; i < MAXN; ++i)Log[i] = (i&(i-1))?Log[i-1]: Log[i-1]+1;}void initRMQ(){for(int i = 1; i <= len; ++i)table[0][i] = height[i];for(int i = 1; (1 << i) <= len; ++i)for(int j = 1; j+(1 << i)-1 <= len; ++j)table[i][j] = min(table[i-1][j], table[i-1][j+(1 << (i-1))]);}int lcp(int a, int b){a = rank[a];b = rank[b];if(a > b) swap(a, b);++a;int temp = Log[b-a+1];return min(table[temp][a], table[temp][b-(1 << temp)+1]);}*/}sa;int len1, len2;inline int idx(int ind){return ind <= len1? 0: 1;}LL ans;int I;int cnt[MAXN][2];struct STACK{LL arr[MAXN][3];int top;void init(int ind){top = 0;arr[top][0] = 0;        //arr[top][2]与前面的串1可以形成多少相同串arr[top][1] = 0;//arr[top][2]与前面的串2可以形成多少相同串arr[top][2] = ind;}void push(int id, int ind){while(top && sa.height[ind] <= sa.height[arr[top][2]])--top;LL t[2];t[0] = arr[top][0]+(cnt[ind-1][0]-cnt[arr[top][2]-1][0])*(sa.height[ind]-K+1);t[1] = arr[top][1]+(cnt[ind-1][1]-cnt[arr[top][2]-1][1])*(sa.height[ind]-K+1);ans += t[!id];++top;arr[top][0] = t[0];arr[top][1] = t[1];arr[top][2] = ind;}}st;int main(){while(scanf("%d", &K), K){scanf("%s", sa.S);len1 = strlen(sa.S);sa.S[len1] = '*';scanf("%s", sa.S+len1+1);len2 = strlen(sa.S);sa.init(len2);sa.getHeight();ans = 0;for(I = 0; I <= len2; ){if(sa.sa[I]+K > len2){++I;continue;}cnt[I-1][0] = cnt[I-1][1] = 0;int id = idx(sa.sa[I]);cnt[I][id] = 1;cnt[I][!id] = 0;st.init(I);int j = I+1;while(j <= len2 && sa.height[j] >= K){id = idx(sa.sa[j]);cnt[j][id] = cnt[j-1][id]+1;cnt[j][!id] = cnt[j-1][!id];st.push(id, j);++j;}I = j;}printf("%I64d\n", ans);}return 0;}