BZOJ 3230 相似子串|后缀数组|RMQ

来源：互联网发布：java开源oa办公系统编辑：程序博客网时间：2024/06/05 12:41

良心提示。
f=两子串的最长公共前缀后缀长度平方和的最大值。
因此开2个后缀数组分别处理前后缀。
而且得开long long..

对于每个子串，其排名与后缀的排名有关，因此对于查询的子串我们找到产生该子串的后缀，而我们可以通过计算前i个后缀产生了多少个子串，并查询之得（STL大法好）。于是RMQ解决之。

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#include <cstdio>#include <algorithm>using namespace std;#define rep(i,j,k) for(int i=j;i<k;i++)#define FOR(i,j,k) for(int i=j;i<=k;i++)const int N = 100100, inf = 0x7fffffff;typedef long long ll;ll read() {    ll s = 0, f = 1; char ch = getchar();    for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;    for (; '0' <= ch && ch <= '9'; ch = getchar()) s = s * 10 + ch - '0';    return s * f;}int log[N]; char str[N];struct SuffixArray {    char r[N];    int n, m, num[N], z[N];    int sa[N], rk[N], h[N], st[20][N];    int wa[N], wb[N], wv[N], c[N];    ll s[N];    SuffixArray(char *_r, int _n, int _m) {        rep(i,0,_n) r[i]=_r[i]-'a'+1; r[_n] = 0;        n = _n + 1; m = _m;        da(); calHeight(); init();    }    void sort(int *sa, int *x, int *y, int n, int m) {        rep(i,0,m) c[i] = 0;        rep(i,0,n) c[x[i]]++;        rep(i,1,m) c[i] += c[i-1];        for(int i=n-1;i>=0;i--) sa[--c[x[i]]] = y[i];    }    void da() {        int j, p, *x = wa, *y = wb, *t;        rep(i,0,n) x[i]=r[i],z[i]=i;        sort(sa, x, z, n, m);        for(j=1,p=1;p<n;j*=2,m=p) {            p=0;            rep(i,n-j,n) y[p++] = i;            rep(i,0,n) if(sa[i]>=j) y[p++]=sa[i]-j;            rep(i,0,n) wv[i]=x[y[i]];            sort(sa, wv, y, n, m);            swap(x,y);p=1;x[sa[0]]=0;            rep(i,1,n) x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;        }    }    void calHeight() {        int k = 0;        rep(i,0,n) rk[sa[i]] = i;        rep(i,0,n-1) {            if(k)k--;            for(int j=sa[rk[i]-1];r[i+k]==r[j+k];k++);            h[rk[i]]=k;        }        rep(i,1,n) s[i] = s[i - 1] + (n - 1 - sa[i]) - h[i];    }    void init() {        rep(i,1,n) st[0][i]=h[i];        FOR(i,1,log[n-1]) rep(j,1,n-(1<<i-1))            st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << i - 1)]);    }    int query(int l, int r) {        l = rk[l], r = rk[r];        if (l > r) swap(l, r); l++;        int t = log[r - l + 1];        return min(st[t][l], st[t][r - (1 << t) + 1]);    }};int main() {    int n, m, l, r, id; ll ans = 0, t, l1, r1, l2, r2, *s;    SuffixArray *sa, *sb;    log[0] = -1; rep(i,1,N) log[i] = log[i / 2] + 1;    n = read(); m = read();    scanf("%s", str); sa = new SuffixArray(str, n, 30); s = sa->s;    reverse(str, str + n); sb = new SuffixArray(str, n, 30);    FOR(i,1,m) {        l = read(); r = read(); ans = 0;        if (l > s[n] || r > s[n]) { puts("-1"); continue; }        id = lower_bound(s + 1, s + n + 1, l) - s;        l1 = sa->sa[id]; r1 = l1 + sa->h[id] - 1 + l - s[id - 1];        id = lower_bound(s + 1, s + n + 1, r) - s;        l2 = sa->sa[id]; r2 = l2 + sa->h[id] - 1 + r - s[id - 1];        t = l1 == l2 ? inf : sa->query(l1, l2);        t = min(t, min(r1 - l1 + 1, r2 - l2 + 1)); ans += t * t;        t = r1 == r2 ? inf : sb->query(n - r1 + 1, n - r2 + 1);        t = min(t, min(r1 - l1 + 1, r2 - l2 + 1)); ans += t * t;        printf("%lld\n", ans);    }    return 0;}