hdu 1080 打表+dp(最长公共子序列)
2015-02-04 01:51
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先把题中给的表格打出来,然后dp
dp[i][j] = max ( dp[i-1][j-1]+a[s1[i]][s2[j]] , dp[i-1][j] + a[s1[i][' '] , dp[i][j-1] + a[' '][s2[j]] )
dp[i][j]代表两个序列当前匹配到几位,如果匹配的话转移如上第一种,如果不匹配有两种情况如上后两种情况,匹配空格
比较恶心的是打表打错了,搞了好久,,醉了....................................
弱啊
dp[i][j] = max ( dp[i-1][j-1]+a[s1[i]][s2[j]] , dp[i-1][j] + a[s1[i][' '] , dp[i][j-1] + a[' '][s2[j]] )
dp[i][j]代表两个序列当前匹配到几位,如果匹配的话转移如上第一种,如果不匹配有两种情况如上后两种情况,匹配空格
比较恶心的是打表打错了,搞了好久,,醉了....................................
弱啊
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #define MAX 107 using namespace std; int t,n,m; int dp[MAX][MAX]; int a[MAX][MAX]; char s1[MAX]; char s2[MAX]; int main ( ) { scanf ( "%d" , &t ); a['A'-64]['A'-64]=a['C'-64]['C'-64]=a['G'-64]['G'-64]=a['T'-64]['T'-64]=5; a['A'-64]['C'-64]=a['C'-64]['A'-64]=a['A'-64]['T'-64]=a['T'-64]['A'-64]=-1; a['A'-64]['G'-64]=a['G'-64]['A'-64]=a['G'-64]['T'-64]=a['T'-64]['G'-64]=-2; a['C'-64]['G'-64]=a['G'-64]['C'-64]=-3; a['C'-64]['T'-64]=a['T'-64]['C'-64]=-2; a[0]['A'-64]=a['A'-64][0] = -3; a[0]['C'-64]=a['C'-64][0] = -4; a[0]['G'-64]=a['G'-64][0] = -2; a[0]['T'-64]=a['T'-64][0] = -1; while ( t-- ) { scanf ( "%d" , &n ); scanf ( "%s" , s1+1 ); scanf ( "%d" , &m ); scanf ( "%s" , s2+1 ); memset ( dp , -0x3f , sizeof ( dp ) ); dp[0][0] = 0; for ( int i = 1 ; i <= n ; i++ ) dp[i][0] = dp[i-1][0] + a[s1[i]-64][0]; for ( int i = 1 ; i <= m ; i++ ) dp[0][i] = dp[0][i-1] + a[0][s2[i]-64]; for ( int i = 1 ; i <= n ; i++ ) for ( int j = 1 ; j <= m ; j++ ) { dp[i][j] = max ( dp[i][j] , dp[i-1][j] + a[s1[i]-64][0] ); dp[i][j] = max ( dp[i][j] , dp[i][j-1] + a[0][s2[j]-64] ); dp[i][j] = max ( dp[i][j] , dp[i-1][j-1] + a[s1[i]-64][s2[j]-64] ); } printf ( "%d\n" , dp [m] ); } }
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