2016多校第四场 HDU5769

给你一个字符x和一个串s，让你找s有多少不同的子串包含字符x设s1,s2，当其长度不同或者某位字符不同则不同。

此类题一般都是要枚举一维，然后往后找，每次加前面未出现过的新串。为了方便去重，这道题可以先用后缀数组处理出来sa,height数组。然后从字典序小的开始往大的找。把x出现的位置丢进可变长数组v，可以构成合法串的开始位置就是x的位置id为了与之前的不重复，利用height的性质可以很快知道sa[i−1]与sa[i]的最长公共前缀，id与起位置去最大值就是不重复的合法位置了。

//只是为了贴个后缀数组。写的博客////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e6 + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;int sa[maxn], rak[maxn], height[maxn];int wa[maxn], wb[maxn], wv[maxn], c[maxn];bool cmp(int *r, int a, int b, int l) {    return r[a] == r[b] && r[a + l] == r[b + l];}//r是需要处理的数组，n是r的长度+1,r[len] = 0;void calcSa(int *r, int *sa, int n, int m) {    int i, j, p, *x = wa, *y = wb;    for(i = 0; i < m; i++) c[i] = 0;    for(i = 0; i < n; i++) c[x[i] = r[i]]++;    for(i = 1; i < m; i++) c[i] += c[i-1];    for(i = n - 1; ~i; i--) sa[--c[x[i]]] = i;    for(j = 1, p = 1; p < n; j *= 2, m = p) {        for(p = 0, 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 < n; i++) wv[i] = x[y[i]];        for(i = 0; i < m; i++) c[i] = 0;        for(i = 0; i < n; i++) c[wv[i]]++;        for(i = 1; i < m; i++) c[i] += c[i-1];        for(i = n - 1; i >= 0; i--) sa[--c[wv[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;    }}//r是需要处理的数组，n是r的长度void calcHeight(int *r, int *sa, int n) {    int i, j, p = 0;    for(i = 1; i <= n; i++)        rak[sa[i]] = i;    for(i = 0; i < n; i++) {        j = sa[rak[i] - 1];        while(r[i+p] == r[j+p]) p++;        height[rak[i]] = p;        if(p > 0) p--;    }}char s[maxn];int r[maxn];int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int T;    int cas = 0;    scanf("%d", &T);    while(T--) {        char t[10];        scanf("%s%s", t, s);        int len = strlen(s);        std::vector<int> v;        for(int i = 0; i < len; i++) {            if(s[i] == t[0])                v.push_back(i);            r[i] = s[i] - 'a' + 1;        }        r[len] = 0;        calcSa(r, sa, len + 1, 30);        calcHeight(r, sa, len);        // for(int i = 1; i <= len; i++) {            // printf("height[%d] = %d\n", i, height[i]);        // }        ll ans = 0;        for(int i = 1; i <= len; i++) {            auto iter = lower_bound(ALL(v), sa[i]);            if(iter == v.end()) continue;            int id = *iter;            // printf("i = %d  id = %d\n", i, id);            id = max(id, sa[i] + height[i]);            // printf("id = %d\n", id);            ans += len - id;        }        printf("Case #%d: %lld\n", ++cas, ans);    }    return 0;}