Codeforces 235C

Codeforces 235C

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题目：给定一主串(S)，(n)次询问，每次询问串(t)的所有循环移位串的出现的次数和

做法：建(SAM)，对于询问串(t)，将他复制一份放在后边，在后缀自动机上匹配，如果匹配长度大于(|t|)，就沿着(fa), 找到第一次大于(|t|)的位置，用这个状态的(right)数组更新答案。注意到可能会匹配到重复的状态，所以要对会更新答案的状态去重。

#include <bits/stdc++.h>
typedef long long ll;
const int N = 1000010 << 1;
using namespace std;

struct SAM{
    int n, step[N], fa[N], ch[N][26], right[N], num[N], last, root, cnt;
	char s[N>>1];
	SAM(){last = root = ++cnt; }
	void add(int x){
		int tmp = s[x] - 'a', p = last, np = ++cnt;
		step[last = np] = x, right[np] = 1;
		while(p && !ch[p][tmp]) ch[p][tmp] = np, p = fa[p];
		if(!p) fa[np] = root;
		else{
			int q = ch[p][tmp];
			if(step[q] == step[p] + 1) fa[np] = q;
			else{
				int nq = ++cnt; step[nq] = step[p] + 1;
				memcpy(ch[nq], ch[q], sizeof(ch[q]));
				fa[nq] = fa[q], fa[q] = fa[np] = nq;
				while(ch[p][tmp] == q) ch[p][tmp] = nq, p = fa[p];
			}
		}
	}
    int tmp[N];
	void calright(){
		memset(tmp, 0, sizeof(tmp));
		for(int i = 1; i <= cnt; i++) tmp[step[i]]++;
		for(int i = 1; i <= n; i++) tmp[i] += tmp[i - 1];
		for(int i = cnt; i; i--) num[tmp[step[i]]--] = i;
		for(int i = cnt; i; i--) right[fa[num[i]]] += right[num[i]];
	}
	void run(){
		scanf(" %s",s+1), n = strlen(s + 1);
		for(int i = 1; i <= n; i++) add(i);
		calright();
	}

    char str[N];
    int len;
    set<int> S;
    ll solve() {
        scanf(" %s",str+1), len = strlen(str+1);
        for(int i = 1; i <= len; ++i) str[i+len] = str[i];
        int now = root, tmp = 0; ll ans = 0;
        for(int i = 1; i <= len+len; ++i) {
            int x = str[i]-'a';
            if(ch[now][x]) now = ch[now][x], ++tmp;
            else {
                while(now && !ch[now][x]) now = fa[now];
                if(!now) now = root, tmp = 0;
                else tmp = step[now]+1, now = ch[now][x];
            }
            while(now!=1 && step[fa[now]] >= len) now = fa[now], tmp = step[now];
            if(tmp >= len) S.insert(now);
        }
        for(set<int>::iterator it = S.begin(); it != S.end(); ++it) ans += right[*it];
        S.clear();
        return ans;
    }
} Fe;

int main() {
    Fe.run();
    int q;
    scanf("%d",&q);
    while(q--) {
        printf("%I64d
",Fe.solve());
    }
}