ZOJ1027 POJ1080Human Gene Functions

非常经典的一条DP题，见以下代码

/******************************************************************************* * Author : Neo Fung * Email : [e-mail] * Last modified : 2011-07-12 16:46 * Filename : ZOJ1027_Human_Gene_Functions.cpp * Description : ZOJ1027 POJ1080Human Gene Functions * *****************************************************************************/// ZOJ1027_Human_Gene_Functions.cpp : 定义控制台应用程序的入口点。////#include "stdafx.h"//#include <fstream>#include <stdio.h>#include <iostream>#include <string>#include <vector>#include <map>#include <math.h>#include <algorithm>#include <numeric>#include <functional>using namespace std;int max(int a,int b,int c){int temp;temp = a>b ? a : b;temp = temp>c ? temp : c;return temp;}int main(void){//ifstream cin("data.txt");int ncase;int alen,blen;int i,j,k;int l,m,n;int alaph[128]={0};char astr[101],bstr[101];int match[5][5]={5,-1,-2,-1,-3,-1,5,-3,-2,-4,-2,-3,5,-2,-2,-1,-2,-2,5,-1,-3,-4,-2,-1,0};int matrix[101][101]={0};alaph['A']=0;alaph['C']=1;alaph['G']=2;alaph['T']=3;cin>>ncase;while(ncase){cin>>alen;for(i=1;i<=alen;++i){cin>>astr[i];}cin>>blen;for(i=1;i<=blen;++i)cin>>bstr[i];for(i=1;i<=alen;++i){matrix[i][0] = matrix[i-1][0] + match[alaph[astr[i]]][4];}for(j=1;j<=blen;++j){matrix[0][j] = matrix[0][j-1] + match[4][alaph[bstr[j]]];//matrix[0][j] = matrix[0][j-1] + match[alaph[astr[j]]][4];}for(i=1;i<=alen;++i)for(j=1;j<=blen;++j){/************************************************************************//*  1：i-1已经与j-1匹配 此时i，j的匹配结果为 matrix[i-1][j-1]+(i与j的匹配值)*//*2：i与j-1匹配  此时 j只能与“-”匹配  结果为 matrix[i][j-1]+(j与'-'的匹配值)*//*  3：i-1与j匹配  此时恰好与2相反                                              *//************************************************************************/l = match[alaph[astr[i]]][alaph[bstr[j]]] + matrix[i-1][j-1];m = match[4][alaph[bstr[j]]] + matrix[i][j-1];n = match[alaph[astr[i]]][4] + matrix[i-1][j];matrix[i][j] = max(l,m,n);}cout << matrix[alen][blen]<<endl;--ncase;}return 0;}