USACO :Longest Prefix解题报告

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一维的dp，设dp[i]表示主串S中从i开始的最长的可组成的前缀，dp[1]就是所求，状态转移方程：

{dp[i]=max{dp[j]+j-i}  其中：i 到 j-1 的字串是属于集合P 且 j 要满足 i<j  && j< n &&   j<=i+10（最后这个限制很重要）

USER: li xiaopeng [xpli1]TASK: prefixLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3708 KB]   Test 2: TEST OK [0.011 secs, 3708 KB]   Test 3: TEST OK [0.000 secs, 3708 KB]   Test 4: TEST OK [0.022 secs, 3708 KB]   Test 5: TEST OK [0.065 secs, 3708 KB]   Test 6: TEST OK [0.464 secs, 3724 KB]All tests OK.Your program ('prefix') produced all correct answers!  This is yoursubmission #9 for this problem.  Congratulations!

/*ID: xpli1PROG: prefixLANG: C++*/

#include <iostream>#include <fstream>#include <string>

using namespace std;

#define IN    fin#define OUT    fout#define max(a,b) (((a) > (b)) ? (a) : (b))

ifstream fin ("prefix.in", ios::in);ofstream fout("prefix.out",ios::out);

string              P[201];string    S;int     max_len;int     dp[200001];

int main(){

 string temp;

 int i,j,k,index = 0;

 while(IN >> temp, temp != ".") { P[index++] = temp; }

 while(IN >> temp) { S += temp; }

 int lens = S.length();

 for(i = lens - 1; i >= 0; i--){

  for(j = i + 1; j <= lens && j <= i + 10; j++){

   for(k = 0; k < index; k++){

    if(P[k].length() != j-i) continue;        if(S.compare(i,j-i,P[k]) == 0) {  dp[i] = max(dp[i],dp[j] + j - i); }   }     } }

 OUT << dp[0] << endl;

 return 0;}