建trie树,刚好字符串是反向的,直接在原图上向前搜索就OK了………………
可怜的我竟然用了RK来hash,在test67那里T了……
贴个RK的
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#define LL long long
using namespace std;
const int MAXN = 10050;
const int MOD = 1e9 + 7;
char tmp[MAXN], str[MAXN];
struct Word{
char s[1005];
int len;
};
int n;
vector<Word>w;
int wmod[MAXN * 10];
int smod[1005];
int ymod[1005];
int dp[MAXN];
int main(){
w.clear();
int len;
scanf("%d", &len);
scanf("%s", tmp);
scanf("%d", &n);
ymod[0] = 1;
for(int i = 1; i <= 1000; i++){
ymod[i] = ((LL)ymod[i - 1] * 26) %MOD;
}
for(int i = len - 1; i >= 0; i--){
str[len - i] = tmp[i];
}
str[len + 1] = '\0';
int minlen = 0;
Word tmps;
for(int i = 0; i < n; i++){
scanf("%s", tmps.s);
tmps. len = strlen(tmps.s);
w.push_back(tmps);
minlen = max(minlen, tmps.len);
}
//cout << 0 << endl;
for(int i = 0; i < n; i++){
int val = 0;
int weight = 1;
for(int j = 0; j < w[i].len; j++){
val = ((LL)val + (LL)(w[i].s[j] >= 'A' && w[i].s[j] <= 'Z'? (w[i].s[j] - 'A') : (w[i].s[j] -'a')) * weight % MOD )%MOD;
weight = (LL)weight * 26 % MOD;
}
wmod[i] = val;
}
/*
for(int i = 0; i < n; i++)
cout << wmod[i] << endl;
*/
memset(smod, -1, sizeof(smod));
smod[0] = 0;
memset(dp, -1, sizeof(dp));
dp[0] = 0;
for(int i = 1; i <= len; i++){
for(int k = minlen + 1; k >= 1; k--){
if(smod[k - 1] == -1) continue;
else{
smod[k] = ((LL)smod[k- 1] + (LL)ymod[k - 1] *( str[i] - 'a' ))%MOD;
}
}
for(vector<Word>::iterator k = w.begin(); k != w.end(); ++k){
char ss = k->s[k->len - 1];
ss = (ss >= 'A' && ss <= 'Z' )? ss -'A' +'a' : ss;
if(ss != str[i] ) continue;
else {
if(smod[k ->len] != -1 && wmod[k - w.begin()] == smod[k->len] && dp[i - k->len] >= 0){
dp[i] = k - w.begin();
break;
}
}
}
}
int i = len;
while(i){
printf("%s", w[dp[i]].s);
i = i - w[dp[i]].len;
if(i) printf(" ");
}
printf("\n");
}
trie树简便快捷啊……
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 1E4 + 10;
struct T{
int ch[26];
int END;
T () {
memset(ch,0,sizeof(ch));
END = 0;
}
}t[1000010];
int cnt,n,m,f[maxn],last[maxn];
char x[maxn],y[1010];
vector <char> v[maxn*10];
void Insert(int o,int pos,int len,int num)
{
for (; pos < len; pos++) {
int to;
if (y[pos] >= 'a') to = y[pos]-'a';
else to = y[pos]-'A';
if (!t[o].ch[to]) t[o].ch[to] = ++cnt;
o = t[o].ch[to];
}
t[o].END = num;
}
void solve(int pos)
{
int END = max(1,pos-999);
int o = 0;
for (int i = pos; i >= END; i--) {
o = t[o].ch[x[i]-'a'];
if (!o) break;
if ((f[i-1] || i == 1) && t[o].END) {
last[pos] = i;
f[pos] = t[o].END;
break;
}
}
}
void pri(int pos)
{
if (last[pos] != 1) pri(last[pos]-1);
int len = v[f[pos]].size();
for (int i = 0; i < len; i++) printf("%c",v[f[pos]][i]);
printf(" ");
}
int main()
{
#ifdef YZY
freopen("yzy.txt","r",stdin);
#endif
cin >> n;
scanf("%s",1+x);
cin >> m;
for (int i = 1; i <= m; i++) {
scanf("%s",&y);
int len = strlen(y);
Insert(0,0,len,i);
for (int j = 0; j < len; j++) v[i].push_back(y[j]);
}
for (int i = 1; i <= n; i++)
solve(i);
pri(n);
return 0;
}