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Description
Input
Output
Sample Input
2 a abc b bbb
Sample Output
Case #1: 3 Case #2: 3
Hint
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <map> #include <set> #include <queue> #include <stack> #include <cmath> #include <vector> #include <bitset> #define ll long long #define rep(x,to) for(int x=0;x<(to);x++) #define repn(x,to) for(int x=1;x<=(to);x++) #define clr(x) memset(x,0,sizeof(x)) #define pli pair<ll, int> #define MEMSET(x,v) memset(x,v,sizeof(x)) #define pll pair<ll,ll> #define pb push_back #define MP make_pair using namespace std; const int N = 5e5 + 100; int t1[N], t2[N], c[N]; bool cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a+l] == r[b+l]; } void da(int str[], int sa[], int Rank[], int heigh[], int n, int m) { n++; int i, j, p, *x = t1, *y = t2; for(i = 0; i < m; ++i) c[i] = 0; for(i = 0; i < n; ++i) c[ x[i] = str[i] ]++; for(i = 1; i < m; ++i) c[i] += c[i - 1]; for(i = n - 1; i >= 0; --i) sa[ --c[ x[i] ] ] = i; for(j = 1; j <= n; j <<= 1) { p = 0; for(i = n - j; i < n; ++i) y[p++] = i; for(i = 0; i < n; ++i) if(sa[i] >= j) y[p++] = sa[i] - j; for(i = 0; i < m; ++i) c[i] = 0; for(i = 0; i < n; ++i) c[ x[ y[i] ] ]++; for(i = 1; i < m; ++i) c[i] += c[i - 1]; for(i = n - 1; i >=0; --i) sa[ --c[ x[ y[i] ] ] ] = y[i]; swap(x, y); p = 1; x[ sa[0] ] = 0; for(i = 1; i < n; ++i) x[ sa[i] ] = cmp(y, sa[i - 1], sa[i], j) ? p - 1: p++; if(p >= n) break; m = p; } int k = 0; n--; for(i = 0; i <= n; ++i) Rank[ sa[i] ] = i; for(i = 0; i < n; ++i) { if(k) k--; j = sa[ Rank[i] - 1 ]; while(str[i + k] == str[j + k]) k++; heigh[ Rank[i] ] = k; } } int Rank[N], heigh[N]; char str[N]; int r[N]; int sa[N]; char X[12]; int pre[N]; int pos[N]; int _ = 1; void solve(int n) { memset(pos, 0, sizeof pos); int p = -1; ll sum = 0, num = 0; for(int i = n - 1; i >= 0; --i) { if(str[i] == X[0]) p = i; pos[i] = p; if(pos[i] != -1) sum += (n - pos[i]); } //for(int i = 0; i < n; ++i) printf("%d ", pos[i]); for(int i = 2; i <= n; ++i) { if(pos[ sa[i - 1] ] == -1 || pos[ sa[i] ] == -1) continue; int c1 = pos[ sa[i - 1] ] - sa[i - 1] + 1; int c2 = pos[ sa[i] ] - sa[i] + 1; int h = heigh[i]; if(c1 > h || c2 > h) continue; num += (h - c1 + 1); } printf("Case #%d: %I64d\n", _++, sum - num); } int main() { #ifdef LOCAL freopen("in", "r", stdin); #endif int cas; scanf("%d", &cas); while(cas --) { scanf("%s%s", X, str); int n = strlen(str); for(int i = 0; i < n; ++i) r[i] = str[i]; r[n] = 0; da(r, sa, Rank, heigh, n, 128); solve(n); // for(int i = 0; i < n; ++i) printf("%d ", Rank[i]); puts(""); // for(int i = 1; i <= n; ++i) printf("%d ", sa[i]); puts(""); // for(int i = 2; i <= n; ++i) printf("%d ", heigh[i]); puts(""); } return 0; }
2016多校联合训练4 F - Substring 后缀数组
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原文地址:http://www.cnblogs.com/orchidzjl/p/5814093.html