poj 1080

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题目链接：http://poj.org/problem?id=1080

就是求基因的相似度，这个题是最长公共子序列的变形，状态方程：

dp[i][j]=Max(dp[i-1][j-1]+map[s1[i-1]][s2[j-1]],dp[i-1][j]+map[s1[i-1]][4],dp[i][j-1]+map[4][s2[j-1]])；

//dp[i][j]表示str1的前i个和str2的前j个的值//注意啊啊啊啊:str1[i]是串中第i+1个字符啊#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <map>//#include <iterator>using namespace std;int maxx(int a,int b){    return a > b ? a : b;}int main(){   string str1,str2;   int n,a1,a2;   int m[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}        };   map<char,int> mapp;   mapp['A'] = 0;   mapp['C'] = 1;   mapp['G'] = 2;   mapp['T'] = 3;   mapp['-'] = 4;    int dp[101][101];     cin >> n;   while(n--)   {       memset(dp,0,sizeof(dp));       cin >> a1 >> str1 >> a2 >> str2;       dp[0][0] = 0;       for(int i = 1; i <= a1; i ++)       {           //只有str1，那么str2全补 ‘-’           //str1前i个，str2没有，即dp[i][0]是由dp[i-1][0]加上m[str1中第i个字符]['-'],这里第i个，就是str1[i-1]啊。。。。此处被坑250次           dp[i][0] = dp[i-1][0] + m[ mapp[str1[i-1]] ][4];         }       for(int i = 1; i <= a2; i++)       {           dp[0][i] = dp[0][i-1] + m [4][mapp[ str2[i-1] ]];       }       for(int i = 1; i <= a1; i++)       {           for(int j = 1; j <= a2; j++)           {               dp[i][j] = maxx(dp[i-1][j-1] + m[ mapp[ str1[i-1] ] ][ mapp[ str2[j-1] ] ],maxx(dp[i][j-1] + m[4][ mapp[str2[j-1]] ],dp[i-1][j] + m[ mapp[str1[i-1]] ][4]));               //cout << i <<" "<<j << " " <<dp[i][j]<<endl;           }       }       cout <<dp[a1][a2]<<endl;   }    return 0;}