Codeforces 347D - Lucky Common Subsequence

给出两个长100的字符串a,b。再额外给出一个长100的字符串 virus。询问a和b最长的没有字串是virus的公共子序列。输出这个子序列。

在经典的LCS的dp解法上再加一维，记录当前的最长公共子序列匹配virus的长度。

第三维转移可以用KMP或者AC自动机。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 112,maxLen = 26;

int toid(char c){ return c - 'A'; }
queue<int> Q;

struct acam{
int nex[maxn][maxLen],fail[maxn];
int cnt[maxn];
int _cnt,root;
int newNode(){
memset(nex[_cnt],fail[_cnt] = -1,sizeof(nex[_cnt]));
cnt[_cnt] = 0;
return _cnt++;
}
void init(){
_cnt = 0;
root = newNode();
}
void insert(char *arr){
int st = root;
for(int i=0;arr[i];i++){
int & stx = nex[st][toid(arr[i])];
if(stx == -1) stx = newNode();
st = stx;
}
cnt[st]++;
// update cnt
}
void build(){
while(Q.empty()==false) Q.pop();
fail[root] = root;
int st;
for(int i=0;i<maxLen;i++){
if((st = nex[root][i]) != -1){
fail[st] = root;
Q.push(st);
}
else nex[root][i] = root;
}
while(Q.empty()==false){
st = Q.front(),Q.pop();
for(int i=0;i<maxLen;i++){
if(nex[st][i] != -1){
int fst = fail[st],son = nex[st][i];
while(fst != root && nex[fst][i] == -1)
fst = fail[fst];
fail[son] = nex[fst][i] == -1 ? root : nex[fst][i];
Q.push(son);
}
else{
nex[st][i] = nex[fail[st]][i];
}
}
}
}
void out(){
for(int i = 0; i<_cnt;i++){
printf("id = %d : ",i);
for(int j = 0;j<maxLen;j++){
if(nex[i][j] != 0){
printf("%c -> %d ",j + 'A',nex[i][j]);
}
}
puts("");
}
}
}acam;

char a[maxn],b[maxn],vair[maxn];
string dp[maxn][maxn][maxn];

string smax(string a,string b){
if(a.size() > b.size())
return a;
return b;
}

int main(){
acam.init();
scanf("%s %s %s",a+1,b+1,vair);
acam.insert(vair);
acam.build();
int n = strlen(a+1),m = strlen(b+1);
int vsiz = strlen(vair);
for(int i = 0;i<=n;i++){
for(int j = 0;j <= m;j++){
for(int k = 0; k <= vsiz; k++){
dp[i][j][k] = "";
}
}
}
for(int i = 1;i<=n;i++){
for(int j = 1;j<=m;j++){
for(int k = 0 ; k < vsiz;k++){
dp[i][j][k] = smax(dp[i][j][k],dp[i-1][j][k]);
dp[i][j][k] = smax(dp[i][j][k],dp[i][j-1][k]);
if(a[i] == b[j]){
int x = a[i] - 'A';
int kx = acam.nex[k][x];
dp[i][j][kx] = smax(dp[i][j][kx],dp[i-1][j-1][k]+string(1,a[i]));
}
}
}
}
string ans = "";
for(int i = 0;i < vsiz;i++){
ans = smax(ans,dp
[m][i]);
}
if(ans == "") ans = "0";
printf("%s\n",ans.c_str());
return 0;
}