NEUOJ 720 (字典树+LCA || 二分+哈希)

题目链接:点击这里

Problem: 头哥的烦恼

Time limit: 5s Mem limit: 1000 MB AC/Submission: 16/94 Discuss
Problem Description
头哥,众所周知,作为acm队的主力选手,最近遇到了一些烦恼,那就是CET-4临近了,但是他还有很多单词没记住.

现在头哥有n个单词没记住,(所有的单词长度加起来不超过5e5),他需要区分其中的m对单词,因为两个单词前面有一部分是一样的,所以他只需要记住后面不相等的部分,所以头哥想要知道每对单词最长相等的前缀长度

Input
The first line of the input contains the number of test cases T.
Each test case begins with two integers n and m, indicate the number of words and queries.
Then n lines follow, each line contains a word (consist only of lowercase letters ‘a’-‘z’).
Then m lines follow, each line contains two integer u, v., meaning Brother want to query the word u and word v. (u may equal to v)

Output
For each test case, output one line containing “Case #x: ”, x means the case number (starting from 1).
Then next m lines, each line just print the answer of the query.

Sample Input
2
3 1
a
aa
aaa
1 3
3 2
ab
abc
bca
1 2
2 3

Sample Output
Case #1:
1
Case #2:
2
0

题意:求两个字符串的最长公共前缀长度.

把所有的字符串加入字典树,然后每次求两个字符串结尾节点的LCA深度就好了.

也可以处理好每一个字符串的哈希序列,然后直接在这个哈希序列上二分.

字典树+LCA

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <map>#include <queue>#include <string>#include <time.h>using namespace std;#define maxn 511111struct node {    int next[26];}tree[maxn];int n, m, cnt, root;char a[maxn];struct E {    int v, next;} edge[maxn<<4];int head[maxn], tot;//LCAint fa[maxn][22], deep[maxn];//节点i第2^j个祖先 深度int DEG;bool vis[maxn];void bfs (int root) {    DEG = 20;    queue <int> q;    while (!q.empty ()) q.pop ();    deep[root] = 0;    fa[root][0] = root;    q.push (root);    memset (vis, 0, sizeof vis); vis[root] = 1;    while (!q.empty ()) {        int u = q.front (); q.pop ();        for (int i = 1; i < DEG; i++) {            fa[u][i] = fa[fa[u][i-1]][i-1];        }        for (int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].v;            if (vis[v]) continue;            deep[v] = deep[u]+1;            fa[v][0] = u;            q.push (v);            vis[v] = 1;        }    }}int LCA (int u, int v) {    if (deep[u] > deep[v])        swap (u, v);    int hu = deep[u], hv = deep[v];    int tu = u, tv = v;    for (int det = hv-hu, i = 0; det; det >>= 1, i++) if (det&1) {        tv = fa[tv][i];    }    if (tu == tv)        return tu;    for (int i = DEG-1; i >= 0; i--) {        if (fa[tu][i] == fa[tv][i])            continue;        tu = fa[tu][i];        tv = fa[tv][i];    }    return fa[tu][0];}void add_edge (int u, int v) {    edge[tot].v = v, edge[tot].next = head[u], head[u] = tot++;}void init () {    tot = cnt = 0;    memset (head, -1, sizeof head);    root = 0;    memset (tree[0].next, -1, sizeof tree[0].next);}//字典树int new_node () {    ++cnt;    memset (tree[cnt].next, -1, sizeof tree[cnt].next);    return cnt;}int insert (char *a) {    int len = strlen (a);    int p = root;    for (int i = 0; i < len; i++) {        int id = a[i]-'a';        if (tree[p].next[id] == -1) {            tree[p].next[id] = new_node ();            add_edge (p, tree[p].next[id]);        }        p = tree[p].next[id];    }    return p;}int pos[maxn];//每一个字符串的尾巴在字典树中对应的节点int main () {    int t, kase = 0;    scanf ("%d", &t);    while (t--) {        init ();        scanf ("%d%d", &n, &m);        for (int i = 1; i <= n; i++) {            scanf ("%s", a);            pos[i] = insert (a);        }        bfs (0);        //dfs (0, 0);        printf ("Case #%d:\n", ++kase);        while (m--) {            int u, v;            scanf ("%d%d", &u, &v);            printf ("%d\n", deep[LCA (pos[u], pos[v])]);        }    }    return 0;}

哈希+二分

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <map>#include <queue>#include <cstdlib>#include <string>using namespace std;#define ull unsigned long long#define seed 131#define maxn 511111vector <ull> a[maxn];int n, m;char str[maxn];ull f (char a) {    return a-'a'+1;}int main () {    int t, kase = 0;    scanf ("%d", &t);    while (t--) {        scanf ("%d%d", &n, &m);        printf ("Case #%d:\n", ++kase);        for (int i = 1; i <= n; i++) {            a[i].clear ();            scanf ("%s", str);            int len = strlen (str);            ull hash = 0;            for (int j = 0; j < len; j++) {                hash = hash*seed + f(str[j]);                a[i].push_back (hash);                if (j)                    a[i][j] += a[i][j-1];            }        }        while (m--) {            int u, v;            scanf ("%d%d", &u, &v);            if (a[u][0] != a[v][0]) {                printf ("0\n");                continue;            }            int r = min (a[u].size (), a[v].size ())-1;            int l = 0;            while (r-l > 1) {                int mid = (l+r)>>1;                if (a[u][mid] == a[v][mid]) {                    l = mid;                }                else                     r = mid;            }            printf ("%d\n", (a[u][r] == a[v][r] ? r : l)+1);        }    }    return 0;}