POJ 1080 Human Gene Functions

题意：是求最长公共子序列的一个变型题；此题中的字符只有A,G,C,T四个，再加上'-'；它们之间的不同匹配有对应的不同权值W；而且只有字符相同的匹配的权值为大于0；其他都为负值。此题就是求权值最大的匹配。

f[i][j]: 表示第一个字符串的前 i 个字符和第二个字符串的前 j 个字符的最大权值。

状态方程：

f[i][j]=max(f[i-1][j]+w[a[i]]['-'],f[i][j-1]+w['-'][b[j]],f[i-1][j-1]+w[a[i]][b[j]]);

// Time 16ms; Memory 324K

#include<iostream>
using namespace std;
int max(int a,int b,int c)
{
	a=a>b?a:b;
	return a>c?a:c;
}
int main()
{
	int i,j,t,n,m,f[110][110],w['T'+1]['T'+1];//权值
	char a[110],b[110];
    w['A']['A']=5;  
    w['C']['C']=5;  
    w['G']['G']=5;  
    w['T']['T']=5;  
    w['A']['C']=w['C']['A']=-1;  
    w['A']['G']=w['G']['A']=-2;  
    w['A']['T']=w['T']['A']=-1;  
    w['A']['-']=w['-']['A']=-3;  
    w['C']['G']=w['G']['C']=-3;  
    w['C']['T']=w['T']['C']=-2;  
    w['C']['-']=w['-']['C']=-4;  
    w['G']['T']=w['T']['G']=-2;  
    w['G']['-']=w['-']['G']=-2;  
    w['T']['-']=w['-']['T']=-1;  
	cin>>t;
	while(t--)
	{
		cin>>n>>a;
		cin>>m>>b;
		f[0][0]=0;
		for(i=1;i<=n;i++) f[i][0]=f[i-1][0]+w[a[i-1]]['-'];//初始化
		for(j=1;j<=m;j++) f[0][j]=f[0][j-1]+w['-'][b[j-1]];
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				f[i][j]=max(f[i-1][j]+w[a[i-1]]['-'],f[i][j-1]+w['-'][b[j-1]],f[i-1][j-1]+w[a[i-1]][b[j-1]]);
			}
		}
		cout<<f[n][m]<<endl;
	}
	return 0;
}

补充：软件开发 , C++ ,