Codeforces 346B

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最长公共子序列加强版,要求最长子序列不包含病毒串

在原本的dp[i][j]上新增一维k,表示A串前i个和B串前j个匹配到病毒串第k个字符的最大值

扩展一下多个病毒串的话就用AC自动机把,K换成自动机状态

A[i]=B[j]时,dp[i+1][j+1][k]=dp[i][j][f[k]]+1

其实这种更新方式挺蛋疼的,从[i-1][j-1]更新过来更符合转移方程定义一些

#include<bits/stdc++.h>#include<stdio.h>#include<algorithm>#include<queue>#include<string.h>#include<iostream>#include<math.h>#include<set>#include<map>#include<vector>#include<iomanip>using namespace std;#define ll long long#define pb push_back#define FOR(a) for(int i=1;i<=a;i++)const int inf=0x3f3f3f3f;const int maxn=5e5+9;  struct pos{int x,y,z;}pre[111][111][111],tail;char a[111],b[111],c[111];int dp[111][111][111];//a前i位b前j位匹配到c串k位时的最长长度int f[111];void getfail(char *p,int *f){f[0]=0;f[1]=0;for(int i=1;p[i];i++){int j=f[i];while(j&&p[i]!=p[j])j=f[j];f[i+1]=p[i]==p[j]?j+1:0;}}void dfs(pos t){if(!t.x && !t.y && !t.z)return;pos q=pre[t.x][t.y][t.z];int i=q.x,j=q.y,k=q.z;dfs(q);if(dp[i][j][k]+1==dp[t.x][t.y][t.z])printf("%c",a[i]);}int main(){scanf("%s%s%s",a,b,c);getfail(c,f);memset(dp,0,sizeof dp);int len1=strlen(a),len2=strlen(b),len3=strlen(c);for(int i=0;i<len1;i++){for(int j=0;j<len2;j++){for(int k=0;k<len3;k++){if(dp[i+1][j][k]<dp[i][j][k]){dp[i+1][j][k]=dp[i][j][k];pre[i+1][j][k].x=i;pre[i+1][j][k].y=j;pre[i+1][j][k].z=k;}if(dp[i][j+1][k]<dp[i][j][k]){dp[i][j+1][k]=dp[i][j][k];pre[i][j+1][k].x=i;pre[i][j+1][k].y=j;pre[i][j+1][k].z=k;}if(a[i]==b[j]){int w=k;while(w&&a[i]!=c[w])w=f[w];if(a[i]==c[w])w++;else w=0;if(dp[i+1][j+1][w]<dp[i][j][k]+1){dp[i+1][j+1][w]=dp[i][j][k]+1;pre[i+1][j+1][w].x=i;pre[i+1][j+1][w].y=j;pre[i+1][j+1][w].z=k;}}}}}/*--[test!]--for(int i=0;i<=len1;i++){for(int j=0;j<=len2;j++){cout<<dp[i][j][0]<<" ";}puts("");}*/int maxx=0;//int i=len1,j=len2;for(int i=0;i<=len1;i++)for(int j=0;j<=len2;j++)for(int k=0;k<len3;k++){if(dp[i][j][k]>maxx){maxx=dp[i][j][k];tail.x=i;tail.y=j;tail.z=k;}}//cout<<maxx<<endl;if(maxx==0)return 0*printf("0\n");dfs(tail);}