poj 1080

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2015/2/13

说不出来是什么dp。。

问在给定矩阵条件的情况下两字符串的最优匹配程度。

那就类似最长公共子序列了...（虽然这样说没有一点科学依据但是你砍我啊！）

字符串s1 和字符串s2

for-> i s1

for->j s2

去遍历匹配，需要注意的是两者都可以匹配这个玩意 ‘ - ’

所以在匹配的时候需要判断下（s1[ i ] - ' - ' 和 ' - ' - s2[ j ] 这样连接时不同的情况看过题目的就明白了）

#include<map>#include<queue>#include<stack>#include<cmath>#include<vector>#include<climits>#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;  #define mod 10007#define lson pos<<1,l,mid#define sc(n) scanf("%d",&n)#define rson pos<<1|1,mid+1,r#define pr(n) printf("%d\n",n)#define met(n,m) memset(n, m, sizeof(n))#define F(x,y,i) for(int i = x;i > y; i--)#define f(x,y,i) for(int i = x;i < y; i++)#define ff(x,y,i) for(int i = x;i <= y; i++)#define FF(x,y,i) for(int i = x;i >= y; i--) const int N=100500;const int inf = INT_MAX;int Max(int a,int b){return a>b?a:b;}int Min(int a,int b){return  a<b?a:b;}int len1,len2;char s1[1005],s2[1005];int tp[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> mp; int dp[105][105];int main()  {      int n, m, tot, x;        mp['A']= 0, mp['C']=1, mp['G']=2, mp['T']=3, mp['-'] = 4;    scanf("%d",&tot);    while(tot--)    {    scanf("%d%s",&len1,s1+1);    scanf("%d%s",&len2,s2+1);dp[0][0] = 0;f(1,len1+1,i)dp[i][0] = tp[mp[s1[i]]][mp['-']] + dp[i-1][0];            f(1,len2+1,i)    dp[0][i] = tp[mp['-']][mp[s2[i]]]+dp[0][i-1];        f(1,len1+1,i)    f(1,len2+1,j)    {    dp[i][j] = dp[i-1][j-1] + tp[mp[s1[i]]][mp[s2[j]]];    dp[i][j] = Max(dp[i][j] , dp[i-1][j] + tp[mp[s1[i]]][mp['-']]);    dp[i][j] = Max(dp[i][j] , dp[i][j-1] + tp[mp['-']][mp[s2[j]]]);    }    printf("%d\n",dp[len1][len2]);    }    return 0;  }

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