POJ 1080 Human Gene Functions

題意：是求最長公用子序列的一個變型題；此題中的字元只有A,G,C,T四個，再加上'-'；它們之間的不同匹配有對應的不同權值W；而且只有字元相同的匹配的權值為大於0；其他都為負值。此題就是求權值最大的匹配。

f[i][j]: 表示第一個字串的前 i 個字元和第二個字串的前 j 個字元的最大權值。

狀態方程：

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];//權值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;}