poj 1080 DP

来源：互联网发布：优化调整方案编辑：程序博客网时间：2024/06/09 22:44

dp【i】【j】表示s1前i个与s2前j个匹配获得的最大值

那么可以 i与j匹配或者 i与‘-’匹配或者jyu‘-’匹配

AC代码如下：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;#define MAX 0x3f3f3f3fchar s1[200], s2[200];int length1, length2;int score[200][200];int dp[200][200];int DFS( int pos1, int pos2 ){    if( pos1 == -1 && pos2 == -1 ){        return 0;    }    if( pos1 != -1 && pos2 != -1 && dp[pos1][pos2] != -1 ){        return dp[pos1][pos2];    }    int ans = -MAX;    if( pos1 != -1 )    ans = max( ans, score[(int)s1[pos1]]['-'] + DFS( pos1 - 1, pos2 ) );    if( pos2 != -1 )    ans = max( ans, score['-'][(int)s2[pos2]] + DFS( pos1, pos2 - 1 ) );    if( pos1 != -1 && pos2 != -1 )  ans = max( ans, score[(int)s1[pos1]][(int)s2[pos2]] + DFS( pos1 - 1, pos2 - 1 ) );    return dp[pos1][pos2] = ans;}int main(){    score['A']['A'] = 5;score['A']['C'] = -1;score['A']['G'] = -2;score['A']['T'] = -1;score['A']['-'] = -3;    score['C']['A'] = -1;score['C']['C'] = 5;score['C']['G'] = -3;score['C']['T'] = -2;score['C']['-'] = -4;    score['G']['A'] = -2;score['G']['C'] = -3;score['G']['G'] = 5;score['G']['T'] = -2;score['G']['-'] = -2;    score['T']['A'] = -1;score['T']['C'] = -2;score['T']['G'] = -2;score['T']['T'] = 5;score['T']['-'] = -1;    score['-']['A'] = -3;score['-']['C'] = -4;score['-']['G'] = -2;score['-']['T'] = -1;score['-']['-'] = -0x3f3f3f;    int T;    scanf( "%d", &T );    while( T-- ){        scanf( "%d%s", &length1, s1 );        scanf( "%d%s", &length2, s2 );        memset( dp, -1, sizeof( dp ) );        cout << DFS( length1 - 1, length2 - 1 ) << endl;    }    return 0;}

0 0