UVa 11081

只确定除了状态量, 没想出状态方程, 这道题的难度还是挺大的.下面是别人的解题报告:

http://par.cse.nsysu.edu.tw/~advprog/advprog2008/11081.doc

代码:

#include<stdio.h>
#include<string.h>
#define MAXN 65

int n;
char s1[MAXN], s2[MAXN], s3[MAXN];
int f1[MAXN][MAXN][MAXN], f2[MAXN][MAXN][MAXN], f[MAXN][MAXN][MAXN];

void solve()
{
    int len1 = strlen(s1+1);
    int len2 = strlen(s2+1);
    int len3 = strlen(s3+1);
    memset(f1,0,sizeof(f1));
    memset(f2,0,sizeof(f2));
    memset(f,0,sizeof(f));
    for(int i = 0; i <= len1; i ++)
        for(int j = 0; j <= len2; j ++)
            f[i][j][0] = 1, f1[i][j][0] = 1, f2[i][j][0] = 1;
        for(int k = 1; k <= len3; k ++)
        {
            for(int i = 0; i <= len1; i ++)
            {
                for(int j = 0; j <= len2; j ++)
                {
                    if(i)//第一串不为空 
                    {
                        f1[i][j][k] = f1[i-1][j][k];
                        if(s1[i] == s3[k]) 
                            f1[i][j][k] += f[i-1][j][k-1];
                            f1[i][j][k] %= 10007;
                    }
                    if(j)//第二串不为空
                    {
                        f2[i][j][k] = f2[i][j-1][k];
                        if(s2[j] == s3[k])
                            f2[i][j][k] += f[i][j-1][k-1];
                            f2[i][j][k] %= 10007;
                    } 
                    f[i][j][k] = (f1[i][j][k] + f2[i][j][k])%10007;
                }
            }
        }
    printf("%d\n",f[len1][len2][len3]);
}
void init()
{
    while(~scanf("%d",&n))
    while(n --)
    {
        scanf("%s%s%s", s1+1,s2+1,s3+1);
        solve();
    }
}
int main()
{
    init();
    return 0;
}