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Senior‘s String

2
a
b
aa
ab

1
2

注：DP  d(i, j)表示最长公共子串长度  f(i, j)表示长度为d(i,j)的子串存在于Y中的数目

递推公式：很简单，选择一个条件，然后分析情况

这里选择取不取X(i)

I. if d(i, j) == d(i-1, j)   then f(i, j) += f(i-1, j);

II.if d(i, j) == d(i-1, p-1) + 1 | X(i)==Y(p) && p<=j   then f(i, j) += f(i-1, p-1).

Node: Initial Conditions--d(0,i) = d(i, 0) = 0, f(0, i) = f(0, i) = 1, f(i, j) = 0;

<pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define maxn 1005
#define MOD 1000000007

char s1[maxn], s2[maxn];
int d[maxn][maxn];
int f[maxn][maxn];
int pos[maxn][27];

void solve()
{
    int len1 = strlen(s1+1);
    int len2 = strlen(s2+1);
    memset(f, 0, sizeof(f));
    for(int i=0; i<=len2; i++)
    {
        d[0][i] = d[i][0] = 0;
        f[0][i] = f[i][0] = 1;
    }

    for(int i=1; i<=len1; i++)
        for(int j=1; j<=len2; j++)
            if(s1[i] == s2[j])
                d[i][j] = d[i-1][j-1] + 1;
            else
                d[i][j] = max(d[i-1][j], d[i][j-1]);

    memset(pos, 0, sizeof(pos));
    for(int i=1; i<=len2; i++)
    {
        for(int j=0; j<26; j++)
            pos[i][j] = pos[i-1][j];
        pos[i][s2[i]-'a'] = i;
    }

    f[0][0] = 1;
    for(int i=1; i<=len1; i++)
        for(int j=1; j<=len2; j++)
        {
            if(d[i][j] == 0)
            {
                f[i][j] = 1;
                continue;
            }
            if(d[i][j] == d[i-1][j])
                f[i][j] = (f[i][j] + f[i-1][j]) % MOD;
            int p = pos[j][s1[i]-'a'];
            if(p && d[i][j] == d[i-1][p-1] + 1)
                f[i][j] = (f[i][j] + f[i-1][p-1]) % MOD;
        }
        printf("%d\n", f[len1][len2]);
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int T;
    cin>>T;
    while(T--)
    {
        scanf("%s", s1+1);
        scanf("%s", s2+1);
        solve();
    }
    return 0;
}

BestCoders

标签：acm   best   

原文地址：http://blog.csdn.net/dojintian/article/details/46928605