UVa 10723 Cyborg Genes (LCS, DP)

题意：给定两行字符串，让你找出一个最短的序列，使得这两个字符串是它的子串，并且求出有多少种。

析：这个题和LCS很像，我们就可以利用这个思想，首先是求最短的长度，不就是两个字符串长度之和再减去公共的么。那么有多少种呢？

同样也是分两种情况讨论，如果s1[i-1] == s2[j-1] 那么种类数肯定和 ans[i-1][j-1]一样了，没有变化，再就是如果不相等怎么算呢？

难道也是ans[i][j] = Max(ans[i-1][j], ans[i][j-1])吗，其实并不是，如果两种方法数相等呢？也就是说从ans[i-1][j]能得到答案，也能从ans[i][j-1]得到答案，

所以要加起来。再就是要注意的是，字符串可能为空串，用gets输入。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s1[maxn], s2[maxn];
int dp[maxn][maxn];
LL ans[maxn][maxn];

int main(){
    int T;  cin >> T;
    getchar();
    for(int kase = 1; kase <= T; ++kase){
        gets(s1+1);
        gets(s2+1);
        int len1 = strlen(s1+1);
        int len2 = strlen(s2+1);
        for(int i = 0; i <= len1; ++i)  ans[i][0] = 1;
        for(int i = 0; i <= len2; ++i)  ans[0][i] = 1;

        for(int i = 1; i <= len1; ++i){
            for(int j = 1; j <= len2; ++j){
                if(s1[i] == s2[j]){
                    dp[i][j] = dp[i-1][j-1] + 1;
                    ans[i][j] = ans[i-1][j-1];
                }
                else if(dp[i-1][j] > dp[i][j-1]){
                    dp[i][j] = dp[i-1][j];
                    ans[i][j] = ans[i-1][j];
                }
                else if(dp[i-1][j] < dp[i][j-1]){
                    dp[i][j] = dp[i][j-1];
                    ans[i][j] = ans[i][j-1];
                }
                else{
                    dp[i][j] = dp[i-1][j];
                    ans[i][j] = ans[i][j-1] + ans[i-1][j];
                }
            }
        }
        printf("Case #%d: %d %lld
", kase, len1+len2-dp[len1][len2], ans[len1][len2]);
    }
    return 0;
}