LOJ#6041 事情的相似度

题意：求字符串[l, r]这些前缀的两两之间最长的最长公共后缀。

把fail树建出来，就是编号在一段区间的最深lca。

套用我之前口胡的莫队做法可以做到O(n√n)，光荣70分。

而且我还非常SB的把原串，fail树，链表上的节点搞混了......

这个链表是那些关键节点按照DFS序串起来的。莫队的时候注意严格按照顺序来搞...

/**
 * There is no end though there is a start in space. ---Infinity.
 * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite.
 * Only the person who was wisdom can read the most foolish one from the history.
 * The fish that lives in the sea doesn't know the world in the land.
 * It also ruins and goes if they have wisdom.
 * It is funnier that man exceeds the speed of light than fish start living in the land.
 * It can be said that this is an final ultimatum from the god to the people who can fight.
 *
 * Steins;Gate
 */

#include <bits/stdc++.h>

const int N = 200010;

struct Edge {
    int nex, v;
}edge[N << 1]; int tp;

int tot = 1, last = 1, tr[N][2], fail[N], len[N], e[N];
int d[N], ST[N << 2][20], pw[N << 2], num2, pos2[N], lc[N], rc[N];
int n, m, id[N], T, fr[N], pre[N], nex[N], Ans[N], id2[N], num;
char str[N];

struct Ask {
    int l, r, id;
    inline bool operator < (const Ask &w) const {
        if(fr[l] != fr[w.l]) return l < w.l;
        return r > w.r;
    }
}ask[N];

namespace BlockV {
    int cnt[N], a[N];
    inline void clear() {
        memset(cnt + 1, 0, fr[n] * sizeof(int));
        memset(a, 0, sizeof(a));
        return;
    }
    inline void insert(int x, int v) {
        a[x] += v;
        cnt[fr[x]] += v;
        //printf("Block add %d %d 
", x, v);
        return;
    }
    inline int getMax() {
        for(int i = fr[n]; i >= 0; i--) {
            if(cnt[i]) {
                for(int j = rc[i]; j >= lc[i]; j--) {
                    if(a[j]) return j;
                }
                return -1;
            }
        }
        return -1;
    }
}

inline bool cmp(const int &a, const int &b) {
    return pos2[a] < pos2[b];
}

inline void add(int x, int y) {
    tp++;
    edge[tp].v = y;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

inline void insert(char c) {
    int f = c == '1', p = last, np = ++tot;
    last = np;
    len[np] = len[p] + 1;
    while(p && !tr[p][f]) {
        tr[p][f] = np;
        p = fail[p];
    }
    if(!p) {
        fail[np] = 1;
    }
    else {
        int Q = tr[p][f];
        if(len[Q] == len[p] + 1) {
            fail[np] = Q;
        }
        else {
            int nQ = ++tot;
            len[nQ] = len[p] + 1;
            fail[nQ] = fail[Q];
            fail[Q] = fail[np] = nQ;
            memcpy(tr[nQ], tr[Q], sizeof(tr[Q]));
            while(tr[p][f] == Q) {
                tr[p][f] = nQ;
                p = fail[p];
            }
        }
    }
    return;
}

void DFS_1(int x, int f) {
    pos2[x] = ++num2;
    ST[num2][0] = x;
    d[x] = d[f] + 1;
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(y == f) {
            continue;
        }
        DFS_1(y, x);
        ST[++num2][0] = x;
    }
    return;
}

inline void prework() {
    for(int i = 2; i <= num2; i++) pw[i] = pw[i >> 1] + 1;
    for(int j = 1; j <= pw[num2]; j++) {
        for(int i = 1; i <= num2; i++) {
            if(d[ST[i][j - 1]] < d[ST[i + (1 << (j - 1))][j - 1]])
                ST[i][j] = ST[i][j - 1];
            else
                ST[i][j] = ST[i + (1 << (j - 1))][j - 1];
        }
    }
    return;
}

inline int lca(int x, int y) {
    x = pos2[x], y = pos2[y];
    if(x > y) std::swap(x, y);
    int t = pw[y - x + 1];
    if(d[ST[x][t]] < d[ST[y - (1 << t) + 1][t]]) {
        return ST[x][t];
    }
    else {
        return ST[y - (1 << t) + 1][t];
    }
}

inline void Add(int x) {
    //printf("add x = %d id = %d 
", x, id[x]);
    x = id[x];
    if(nex[x]) pre[nex[x]] = x;
    if(pre[x]) nex[pre[x]] = x;
    if(nex[x] && pre[x]) BlockV::insert(len[lca(nex[x], pre[x])], -1);
    if(nex[x]) {
        BlockV::insert(len[lca(x, nex[x])], 1);
    }
    if(pre[x]) {
        BlockV::insert(len[lca(x, pre[x])], 1);
    }
    return;
}

inline void Del(int x) {
    //printf("del x = %d id = %d 
", x, id[x]);
    x = id[x];
    if(pre[x]) nex[pre[x]] = nex[x];
    if(nex[x]) pre[nex[x]] = pre[x];
    if(nex[x] && pre[x]) {
        BlockV::insert(len[lca(nex[x], pre[x])], 1);
    }
    if(nex[x]) {
        BlockV::insert(len[lca(x, nex[x])], -1);
    }
    if(pre[x]) {
        BlockV::insert(len[lca(x, pre[x])], -1);
    }
    return;
}

int main() {
    scanf("%d%d", &n, &m);
    T = n / sqrt(m);
    //printf("T = %d sqrt(m)= %d 
", T, (int)sqrt(m));
    for(int i = 1; i <= n; i++) {
        fr[i] = (i - 1) / T + 1;
        //printf("fr %d = %d 
", i, fr[i]);
    }
    for(int i = 1; i <= fr[n]; i++) {
        lc[i] = rc[i - 1] + 1;
        rc[i] = lc[i] + T - 1;
        if(i == fr[n]) rc[i] = n;
    }
    scanf("%s", str + 1);
    for(int i = 1; i <= m; i++) {
        scanf("%d%d", &ask[i].l, &ask[i].r);
        ask[i].id = i;
    }
    std::sort(ask + 1, ask + m + 1);
    for(int i = 1; i <= n; i++) {
        insert(str[i]);
        id[i] = id2[i] = last;
    }
    for(int i = 2; i <= tot; i++) {
        add(fail[i], i);
        //printf("add : %d %d 
", fail[i], i);
    }
    DFS_1(1, 0);
    prework();
    /// solve

    std::sort(id2 + 1, id2 + n + 1, cmp);

    int p = 1;
    for(int i = 1; i <= n; i++) {
        p = tr[p][str[i] == '1'];
        if(id[i] != p) {
            printf("ERROR! 
");
        }
    }

    for(int i = 1; i < n; i++) {
        int x = id2[i], y = id2[i + 1];
        //printf("i = %d x = %d y = %d 
", i, x, y);
        nex[x] = y;
        pre[y] = x;
        BlockV::insert(len[lca(x, y)], 1);
    }
    p = 1;
    int l = 1, r = n;
    for(int i = 1; i <= fr[n]; i++) {
        while(r < n) Add(++r);
        while(l < lc[i]) Del(l++);
        while(fr[ask[p].l] == i) {
            while(ask[p].r < r) Del(r--);
            while(l < ask[p].l) Del(l++);
            Ans[ask[p].id] = BlockV::getMax();
            //printf("id = %d 
", ask[p].id);
            while(lc[i] < l) Add(--l);
            p++;
        }
    }
    for(int i = 1; i <= m; i++) {
        printf("%d
", Ans[i]);
    }
    return 0;
}

然后是启发式合并 + 树状数组 + 扫描线代码。

/**
 * There is no end though there is a start in space. ---Infinity.
 * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite.
 * Only the person who was wisdom can read the most foolish one from the history.
 * The fish that lives in the sea doesn't know the world in the land.
 * It also ruins and goes if they have wisdom.
 * It is funnier that man exceeds the speed of light than fish start living in the land.
 * It can be said that this is an final ultimatum from the god to the people who can fight.
 *
 * Steins;Gate
 */

#include <bits/stdc++.h>

const int N = 200010;

struct Edge {
    int nex, v;
}edge[N << 1]; int tp;

struct Ask {
    int l, r, id;
    Ask(int L = 0, int R = 0, int ID = 0) {
        l = L;
        r = R;
        id = ID;
    }
    inline bool operator < (const Ask &w) const {
        return l < w.l;
    }
}ask[N], node[N * 20]; int Cnt;
typedef Ask Node;

char str[N];
int e[N], n, m, tot = 1, last = 1, tr[N][2], fail[N], len[N], id[N], Ans[N], ta[N];
std::set<int> st[N];
std::set<int>::iterator it, it2;

inline void add(int x, int y) {
    tp++;
    edge[tp].v = y;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

inline void Add(int x, int y) {
    //printf("add %d %d 
", x, y);
    for(int i = x; i <= n; i += i & (-i)) {
        ta[i] = std::max(ta[i], y);
    }
    return;
}

inline int Ask(int x) {
    //printf("ask %d 
", x);
    int ans = 0;
    for(int i = x; i >= 1; i -= i & (-i)) {
        ans = std::max(ans, ta[i]);
    }
    return ans;
}

inline void insert(char c) {
    int f = c == '1', p = last, np = ++tot;
    last = np;
    len[np] = len[p] + 1;
    while(p && !tr[p][f]) {
        tr[p][f] = np;
        p = fail[p];
    }
    if(!p) {
        fail[np] = 1;
    }
    else {
        int Q = tr[p][f];
        if(len[Q] == len[p] + 1) {
            fail[np] = Q;
        }
        else {
            int nQ = ++tot;
            len[nQ] = len[p] + 1;
            fail[nQ] = fail[Q];
            fail[Q] = fail[np] = nQ;
            memcpy(tr[nQ], tr[Q], sizeof(tr[Q]));
            while(tr[p][f] == Q) {
                tr[p][f] = nQ;
                p = fail[p];
            }
        }
    }
    return;
}

void DFS_1(int x) {
    if(id[x]) {
        st[x].insert(id[x]);
        //printf("id  %d =  %d 
", x, id[x]);
    }
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        DFS_1(y);
        /// merge
        if(st[x].size() < st[y].size()) {
            std::swap(st[x], st[y]);
        }
        for(it = st[y].begin(); it != st[y].end(); it++) {
            int a = *it;
            it2 = st[x].lower_bound(a);
            if(it2 != st[x].end()) {
                int b = *it2;
                /// node a b
                if(a > b) std::swap(a, b);
                node[++Cnt] = Node(a, b, len[x]);
                //printf("Node : %d %d %d 
", a, b, len[x]);
            }
            if(it2 != st[x].begin()) {
                it2--;
                int b = *it2;
                /// node a b
                if(a > b) std::swap(a, b);
                node[++Cnt] = Node(a, b, len[x]);
                //printf("Node : %d %d %d 
", a, b, len[x]);
            }
        }
        for(it = st[y].begin(); it != st[y].end(); it++) {
            st[x].insert(*it);
        }
        st[y].clear();
    }
    return;
}

int main() {
    scanf("%d%d", &n, &m);
    scanf("%s", str + 1);
    for(int i = 1; i <= m; i++) {
        scanf("%d%d", &ask[i].l, &ask[i].r);
        ask[i].id = i;
    }
    for(int i = 1; i <= n; i++) {
        insert(str[i]);
        id[last] = i;
    }
    for(int i = 2; i <= tot; i++) {
        add(fail[i], i);
    }
    DFS_1(1);

    std::sort(ask + 1, ask + m + 1);
    std::sort(node + 1, node + Cnt + 1);

    int p1 = Cnt, p2 = m;
    for(int i = n; i >= 1; i--) {
        //printf("i = %d 
", i);
        while(node[p1].l == i) {
            Add(node[p1].r, node[p1].id);
            p1--;
        }
        while(ask[p2].l == i) {
            Ans[ask[p2].id] = Ask(ask[p2].r);
            p2--;
        }
    }
    for(int i = 1; i <= m; i++) {
        printf("%d 
", Ans[i]);
    }
    return 0;
}