HDU 5384 Danganronpa

时间：2015-08-13 23:45:08 阅读：221 评论：0 收藏：0 [点我收藏+]

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Danganronpa

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 241 Accepted Submission(s): 137

Problem Description

Danganronpa is a video game franchise created and developed by Spike Chunsoft, the series‘ name is compounded from the Japanese words for "bullet" (dangan) and "refutation" (ronpa).

Now, Stilwell is playing this game. There are

n verbal evidences, and Stilwell has

m "bullets". Stilwell will use these bullets to shoot every verbal evidence.

Verbal evidences will be described as some strings

Ai, and bullets are some strings

Bj. The damage to verbal evidence

Ai from the bullet

Bj is

f(Ai,Bj).

f (A, B) = \sum i = 1 | A | ? | B | + 1 [A [i . . . i + | B | ? 1] = B]

In other words,

f(A,B) is equal to the times that string

B appears as a substring in string

A.
For example:

f(ababa,ab)=2,

f(ccccc,cc)=4

Stilwell wants to calculate the total damage of each verbal evidence

Ai after shooting all

m bullets

Bj, in other words is

∑mj=1f(Ai,Bj).

Input

The first line of the input contains a single number

T, the number of test cases.
For each test case, the first line contains two integers

m.
Next

n lines, each line contains a string

Ai, describing a verbal evidence.
Next

m lines, each line contains a string

Bj, describing a bullet.

T≤10
For each test case,

n,m≤105,

1≤|Ai|,|Bj|≤104,

∑|Ai|≤105,

∑|Bj|≤105
For all test case,

∑|Ai|≤6?105,

∑|Bj|≤6?105,

Ai and

Bj consist of only lowercase English letters

Output

For each test case, output

n lines, each line contains a integer describing the total damage of

Ai from all

m bullets,

∑mj=1f(Ai,Bj).

Sample Input

1
5 6
orz
sto
kirigiri
danganronpa
ooooo
o
kyouko
dangan
ronpa
ooooo
ooooo

Sample Output

Source

2015 Multi-University Training Contest 8

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<edl
typedef long long ll;
const int inf = 0x3f3f3f3f;

const int M=6e5+7, N=1e5+8;
int ch[M][26], ed[M];
int pos[N];
int L, root = 0;
int fail[M];
char str[M], buf[M];
int newnode()
{
    memset(ch[L], -1, sizeof ch[L]);
    ed[L++] = 0;
    return L - 1;
}
void AC_init() { L = 0; newnode(); }
void insert(char s[])
{
    int n = strlen(s);
    int u = 0;
    for (int i=0;i<n;i++) {
        int &v = ch[u][s[i] - 'a'];
        if (v == -1)
            v = newnode();
        u = v;
    }
    ed[u] ++;
}
void AC_build()
{
    queue<int> q;
    for (int i=0;i<26;i++) {
        int &v = ch[0][i];
        if (v != -1) {
            q.push(v);
            fail[v] = 0;
        } else v = 0;
    }
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i=0;i<26;i++) {
            int &v = ch[u][i];
            if (v==-1) {
                v = ch[fail[u]][i];
            } else {
                fail[v] = ch[fail[u]][i];
                q.push(v);
            }
        }
    }
}
ll query(char buf[], int len)
{
    ll ret = 0; int u = 0;
    for (int i=0;i<len;i++) {
        u = ch[u][buf[i] - 'a'];
        for (int v=u; v>0; v=fail[v])
            ret += ed[v];
    }
    return ret;
}
int n, m;
int main()
{
    int re, ca= 1; scanf("%d", &re);
    while (re--) {
        scanf("%d%d", &n, &m);
        pos[0] = 0;
        for (int i=0;i<n;i++) {
            scanf("%s", str + pos[i]);
            pos[i+1 ] = strlen(str);
        }
        AC_init();
        for (int i=0;i<m;i++) {
            scanf("%s", buf);
            insert(buf);
        }
        AC_build();
        for (int i=0;i<n;i++) {
            ll ans = 0;
            ans = query(str + pos[i], pos[i+1] - pos[i]);
            printf("%I64d\n", ans);
        }
    }
    return 0;
}

HDU 5384 Danganronpa

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原文地址：http://blog.csdn.net/oilover/article/details/47622913

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