HDU-4681 String 枚举+DP
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4681
题意：给A,B,C三个串，求一个最长的串D，满足D是A和B的subsequence，C是D的substring。。
比赛那天把substing搞成了subsequence,,,sd。。。
挺水的一题，直接枚举C在A和B串中的位置，当然是最短的位置，然后求两遍A和B的最长公共子序列，一个从前往后，另一个从后往前，然后遍历枚举就可以了，O(n^2)..
1 //STATUS:C++_AC_343MS_8164KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=1010; 37 const int INF=0x3f3f3f3f; 38 const LL MOD=1000000007,STA=8000010; 39 const LL LNF=1LL<<55; 40 const double EPS=1e-9; 41 const double OO=1e50; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 char A[N],B[N],C[N]; 59 int da[N][2],db[N][2],f1[N][N],f2[N][N]; 60 int T; 61 62 void getd(char a[],char c[],int d[][2],int &cnt) 63 { 64 cnt=0; 65 int i,j,k,len=strlen(a),lenc=strlen(c); 66 for(i=0;i<len;i++){ 67 if(a[i]!=c[0])continue; 68 for(j=i,k=0;j<len;j++){ 69 if(a[j]==c[k]){ 70 k++; 71 if(k==lenc)break; 72 } 73 } 74 if(k==lenc){ 75 d[cnt][0]=i; 76 d[cnt++][1]=j; 77 } 78 } 79 } 80 81 int main(){ 82 // freopen("in.txt","r",stdin); 83 int i,j,lena,lenb,lenc,ca=1; 84 int cnta,cntb,ans; 85 scanf("%d",&T); 86 while(T--) 87 { 88 scanf("%s%s%s",A,B,C); 89 lena=strlen(A);lenb=strlen(B);lenc=strlen(C); 90 getd(A,C,da,cnta); 91 getd(B,C,db,cntb); 92 93 for(i=0;i<=lena || i<=lenb;i++) 94 f1[i][0]=f1[0][i]=f2[i][lenb]=f2[lena][i]=0; 95 for(i=1;i<=lena;i++){ 96 for(j=1;j<=lenb;j++){ 97 f1[i][j]=Max(f1[i][j-1],f1[i-1][j]); 98 if(A[i-1]==B[j-1])f1[i][j]=Max(f1[i][j],f1[i-1][j-1]+1); 99 } 100 } 101 for(i=lena-1;i>=0;i--){ 102 for(j=lenb-1;j>=0;j--){ 103 f2[i][j]=Max(f2[i+1][j],f2[i][j+1]); 104 if(A[i]==B[j])f2[i][j]=Max(f2[i][j],f2[i+1][j+1]+1); 105 } 106 } 107 ans=0; 108 for(i=0;i<cnta;i++){ 109 for(j=0;j<cntb;j++){ 110 ans=Max(ans,f1[da[i][0]][db[j][0]]+f2[da[i][1]+1][db[j][1]+1]); 111 } 112 } 113 if(!cnta || !cntb)ans=-lenc; 114 115 printf("Case #%d: %d\n",ca++,ans+lenc); 116 } 117 return 0; 118 }
