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10366.cpp
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10366.cpp
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#include<bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define pb push_back
typedef long long ll;
// 1. 각 pattern에 대해 insert 함수 실행
// 2. Aho 함수 실행(Fail 함수는 Aho함수가 call함)
// 3. 현재 코드는 패턴의 등장 횟수를 계산하는 코드. 상황에 맞게 Aho, c2i 함수 수정
class Node {
public:
int pt, fail;
vector<int> next;
Node(int num):next(num) {
pt = 0;
fail = 0;
}
};
class Trie {
public:
int nsz; // node size
vector<Node> node;
int mx; // max node num
int num; // kind of character(ex : a to z -> 26, 0 to 9 -> 10)
Trie(int mx, int num) : mx(mx), num(num) {
node.resize(mx, Node(num));
nsz = 1;
}
int c2i(char c) { // character to int, 0 to num-1
return c-'A';
}
void insert(string& p, int strcost) // ptag : pattern의 tag
{
int cur = 0;
for (auto& c : p) {
int n = c2i(c);
if (node[cur].next[n] == 0) {
node[cur].next[n] = nsz;
cur = nsz++;
}
else cur = node[cur].next[n];
}
node[cur].pt = strcost;
}
void Fail() { // 0 : root
queue<int> q;
node[0].fail = 0;
q.push(0);
while (!q.empty()) {
int cur = q.front();
q.pop();
for (int n = 0; n < num; ++n) {
int child = node[cur].next[n];
if (!child) continue;
if (cur == 0)
node[child].fail = 0;
else {
int t = node[cur].fail;
while (t != 0 && node[t].next[n] == 0) t = node[t].fail;
if (node[t].next[n]) t = node[t].next[n];
node[child].fail = t;
}
if (node[node[child].fail].pt != 0)
node[child].pt += node[node[child].fail].pt;
q.push(child);
}
}
}
int findnxt(int cur, int n){
while(cur and node[cur].next[n] == 0) cur = node[cur].fail;
return node[cur].next[n];
}
int Aho(string& s) {
Fail();
int cnt = 0;
int cur = 0;
for (int i = 0; i < s.size(); i++) {
int n = c2i(s[i]);
while (cur and node[cur].next[n] == 0) cur = node[cur].fail;
if (node[cur].next[n]) cur = node[cur].next[n];
if (node[cur].pt != -1) cnt += node[cur].pt;
}
return cnt;
}
};
int main(void){
ios::sync_with_stdio(0);
cin.tie(0);
int T;
cin >> T;
for(int Case = 1; Case <= T; Case++){
int N,M,B;
cin >> N >> M >> B;
int cost[26];
Trie TR(10000,26);
fill(cost,cost+26,-1);
for(int i = 0; i < N; i++){
string s;
int c;
cin >> s >> c;
cost[s[0]-'A'] = c;
}
for(int i = 0; i < M; i++){
string s;
int c;
cin >> s >> c;
TR.insert(s,c);
}
TR.Fail();
int D[10000][201] = {};
memset(D,0xff,sizeof(D));
int mx = 0;
D[0][0]=0;
for(int i = 0; i <= B; i++){
for(int j = 0; j < TR.nsz; j++){
if(D[j][i]==-1) continue;
mx = max(mx,D[j][i]);
// printf("D[%d][%d] : %d\n",j,i,D[j][i]);
for(int n = 0; n < 26; n++){
if(cost[n] == -1 or i+cost[n] > B) continue;
int nxt = TR.findnxt(j,n);
if(nxt == 0) continue;
// printf("n : %d\n",n);
D[nxt][i+cost[n]] = max(D[nxt][i+cost[n]],D[j][i]+TR.node[nxt].pt);
}
}
}
cout << "Case #" << Case << ": " << mx << '\n';
}
}