Uva 10405 Longest Common Subsequence
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Difficulty: easy
Algorithms: Dynamic Programming
Source: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=16&problem=1346&mosmsg=Submission+received+with+ID+8357683
Solution Description: 这是一个经典的动态规划问题。设两个序列Xi,Yj,d[i,j]表示Xi和Yj的一个LCS的长度,如i= 0 或j = 0,其中一个序列的长度为0,因此LCS的长度为0,状态转移方程为:if xi = yj, d[i, j] = d[i - 1, j - 1] + 1;else d[i, j] = max{d[i - 1, j] d[i, j - 1]};注意在初始化时,若i = 0 或j = 0,d[i, j] = 0,然后利用递推计算或记忆化搜索就可以求出,其中d[length(Xi), length(Yj)]就是结果。注意本题不可用一维的滚动数组来实现,因为d[i - 1, j - 1]会被d[i, j - 1]覆盖掉。
递推计算:
memset(d, 0, sizeof(d));//初始化
for(i = 1; i <= m; i++)
for(j = 1; j <= n; j++){
if(x[i - 1] == y[j - 1])
d[i][j] = d[i - 1][j - 1] + 1;
else if(d[i][j - 1] > d[i - 1][j])
d[i][j] = d[i][j - 1];
else d[i][j] = d[i - 1][j];
}
记忆化搜索:
memset(d, -1, sizeof(d));//注意初始化边界
for(i = 0; i <= m; i++)
d[i][0] = 0;
for(i = 0; i <= n; i++)
d[0][i] = 0;
int df(int i, int j)
{
if(d[i][j] >= 0)
return d[i][j];
if(x[i - 1] == y[j - 1])
d[i][j] = df(i - 1, j - 1) + 1;
else d[i][j] = df(i, j - 1) >? df(i - 1, j);
return d[i][j];
}
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