POJ 1080（最长公共子序列）

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题目链接：http://poj.org/problem?id=1080

一道最长公共子序列变形的题目。

关于那个表格，没想到很好的处理方式，直接打表了。

最长公共子序列的状态转移方程：

if(a[i]==b[i]) dp[i][j]=dp[i-1][j-1]+1;

else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

这道题需要做一下变形：

s1取字母，s2取字母:dp[i-1][j-1]+f[s1[i]][s2[j]]

s1取'-'，s2取字母:dp[i][j-1]+f['-'][s2[j]]

s1取字母，s2取'-':dp[i-1][j]+f[s1[i]]['-']

dp[i][j]等于三者的最大值。

在求解之前，需要初始化：

dp[0][0]=0;

dp[i][0]=dp[i-1][0]+f[s1[i]]['-'];

dp[0][j]=dp[0][j-1]+f['-'][s2[j]];

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;const int maxn=105;int T,n,m;char s1[maxn],s2[maxn];int dp[maxn][maxn],f[maxn][maxn];void work(){f['A']['A']=5;f['C']['A']=-1;f['G']['A']=-2;f['T']['A']=-1;f['-']['A']=-3;f['A']['C']=-1;f['C']['C']=5;f['G']['C']=-3;f['T']['C']=-2;f['-']['C']=-4;f['A']['G']=-2;f['C']['G']=-3;f['G']['G']=5;f['T']['G']=-2;f['-']['G']=-2;f['A']['T']=-1;f['C']['T']=-2;f['G']['T']=-2;f['T']['T']=5;f['-']['T']=-1;f['A']['-']=-3;f['C']['-']=-4;f['G']['-']=-2;f['T']['-']=-1;f['-']['-']=INF;}void initData(){dp[0][0]=0;for(int i=1;i<=n;i++)dp[i][0]=dp[i-1][0]+f[s1[i]]['-'];for(int j=1;j<=m;j++)dp[0][j]=dp[0][j-1]+f['-'][s2[j]];}int main(){#ifndef ONLINE_JUDGE    freopen("test.in","r",stdin);    freopen("test.out","w",stdout);#endif    work();    scanf("%d",&T);    while(T--){    scanf("%d%s%d%s",&n,s1+1,&m,s2+1);    initData();    for(int i=1;i<=n;i++){    for(int j=1;j<=m;j++){    int tmp1=dp[i-1][j-1]+f[s1[i]][s2[j]];    int tmp2=dp[i-1][j]+f[s1[i]]['-'];    int tmp3=dp[i][j-1]+f['-'][s2[j]];    dp[i][j]=max(max(tmp1,tmp2),tmp3);    }    }    printf("%d\n",dp[n][m]);    }return 0;}

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