POJ 1080 Human Gene Functions DP

传送门：POJ1080

以下转自：http://blog.csdn.net/lyy289065406/article/details/6648156

设dp[i][j]为取s1第i个字符，s2第j个字符时的最大分值

则决定dp为最优的情况有三种（score[][]为s1[i]和s2[j]两符号的分数）：

1、  s1取第i个字母，s2取“ - ”： dp[i-1][j]+score[ s1[i-1] ]['-'];

2、  s1取“ - ”，s2取第j个字母：dp[i][j-1]+score['-'][ s2[j-1] ];

3、  s1取第i个字母，s2取第j个字母：dp[i-1][j-1]+score[ s1[i-1] ][ s2[j-1] ];

 即dp[i][j]=max( dp[i-1][j]+score[ s1[i-1] ]['-'],

dp[i][j-1]+score['-'][ s2[j-1] ],

dp[i-1][j-1]+score[ s1[i-1] ][ s2[j-1] ] );

 

注意初始化

不仅仅只有

dp[0][0] = 0

也不仅仅是

dp[0][0] = 0

dp[1][0] = score[ s1[i-1] ]['-']

dp[0][1] = score['-'][ s2[j-1] ]

必须全面考虑到所有情况，

当i=j=0时，dp[i][j]=0

当i=0时，dp[0,j] = dp[0][j-1] + score['-'][ s2[j-1] ]

当j=0时，dp[i,0] = dp[i-1][0] + score[ s1[i-1] ]['-']

侵删。

此题还推荐看一下discuss里第一篇帖子，传送门：http://poj.org/showmessage?message_id=74842

看了都说好。

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#define ll long long
#define pi acos(-1)
#define inf 0x3f3f3f3f
using namespace std;
typedef pair<int,int>P;
int hash[100];
int dp[110][110];
char s1[105],s2[105];
int main()
{
hash['A']=0;hash['C']=1;
hash['G']=2;hash['T']=3;hash['-']=4;
int score[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 t;
scanf("%d",&t);
while(t--)
{
int n1,n2;
memset(dp,0,sizeof(dp));
scanf("%d %s",&n1,s1+1);
scanf("%d %s",&n2,s2+1);
for(int i=1;i<=n1;i++)dp[0][i]=dp[0][i-1]+score[4][hash[s2[i]]];
for(int i=1;i<=n2;i++)dp[i][0]=dp[i-1][0]+score[hash[s1[i]]][4];
for(int i=1;i<=n1;i++)
for(int j=1;j<=n2;j++)
{
dp[i][j]=max(dp[i-1][j-1]+score[hash[s1[i]]][hash[s2[j]]],max(dp[i][j-1]+score[4][hash[s2[j]]],dp[i-1][j]+score[hash[s1[i]]][4]));
}
printf("%d\n",dp[n1][n2]);
}
return 0;
}