UVA 11775 Unique Story（LCS [LIS] + BIT）

题意:

$求两个序列不相同的序列个数, n \leq 1000$

分析:

$- - 我直接正着搞, dp状态很显然, dp[i][j]:=以a[i], b[j]结尾的不相同的序列个数$
$但事实上非常难转移, 存在重复的情况, 搞了昨晚一个晚上也没解决$
$如果反正思考的话, 考虑总子序列数(2^n+2^m-2)减去相同的子序列数, 这个题就变得非常显然和简单了$

---

$O(nmlognlogm)$

$1.\ 考虑dp[i][j]:=以a[i],b[j]结尾的相同的子序列的个数$
$转移只有一种当a[i]==b[j]时, dp[i][j] = \sum dp[x][y](x<i, y<j) + 2$
$这个二维的前缀和, 我们可以用二维树状数组来优化就好了$
$dp[i][j] = sum(i,j-1)+sum(i-1,j)-sum(i-1,j-1)+2, 就是去掉右下角的(i,j)这个点$
$ans_{相同的子序列个数} = sum(n,m)$
$复杂度为O(nmlognlogm)$

代码:

//
//  Created by TaoSama on 2015-11-22
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 10000007;

int n, m, u[N], v[N];
int b[N][N], dp[N][N];

void add(int &x, int y) {
    x = x + y + MOD;
    while(x >= MOD) x -= MOD;
}

void add(int x, int y, int v) {
    for(int i = x; i <= n; i += i & -i)
        for(int j = y; j <= m; j += j & -j)
            add(b[i][j], v);
}

int sum(int x, int y) {
    int ret = 0;
    for(int i = x; i; i -= i & -i)
        for(int j = y; j; j -= j & -j)
            add(ret, b[i][j]);
    return ret;
}

int cnt = 0;
map<string, int> mp;

int ID(string &s) {
    if(mp.count(s)) return mp[s];
    mp[s] = ++cnt;
    return cnt;
}

void handle(int &n, string &s, int *u) {
    n = 0; s += 'A';
    string tmp;
    for(int i = 0; i < s.size(); ++i) {
        if(isalpha(s[i])) {
            if(tmp.size()) {
                u[++n] = ID(tmp);
                tmp.clear();
            }
        }
        tmp += s[i];
    }
}

int two[2005] = {1};

string s, t;

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int T; cin >> T;
    int kase = 0;
    for(int i = 1; i <= 2000; ++i) two[i] = two[i - 1] * 2 % MOD;
    while(T--) {
        cin >> s >> t;
        cnt = 0; mp.clear();
        handle(n, s, u);
        handle(m, t, v);

        memset(dp, 0, sizeof dp);
        memset(b, 0, sizeof b);
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                if(u[i] != v[j]) continue;
                add(dp[i][j], sum(i, j - 1) + sum(i - 1, j)
                    - sum(i - 1, j - 1) + 2);
                add(i, j, dp[i][j]);
            }
        }
        int ans = ((two[n] + two[m] - 2 - sum(n, m)) % MOD + MOD) % MOD;
        printf("Case %d: %d\n", ++kase, ans);
    }
    return 0;
}

---

$O(nm)$

$仔细想想, 其实我们可以把前缀和过程直接放在转移里, 只要换一下状态$
$2.\ 考虑dp[i][j]:=处理到a[i],b[j]的相同的子序列的个数$
$a[i]==b[j], 之前的+之前的加上a[i],b[j]+只有a[i],b[j]$
$dp[i][j] = 2 * dp[i-1][j-1]+2$
$a[i]!=b[j], 把之前所有的加起来, 其实就是去掉(i,j)那个点的和上面的一样$
$dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1]$
$ans_{相同的子序列个数} = dp[n][m]$
$复杂度为O(nm)$

代码:

//
//  Created by TaoSama on 2015-11-22
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 10000007;

int n, m, u[N], v[N];
int b[N][N], dp[N][N];

void add(int &x, int y) {
    x = x + y + MOD;
    while(x >= MOD) x -= MOD;
}

int cnt = 0;
map<string, int> mp;

int ID(string &s) {
    if(mp.count(s)) return mp[s];
    mp[s] = ++cnt;
    return cnt;
}

void handle(int &n, string &s, int *u) {
    n = 0; s += 'A';
    string tmp;
    for(int i = 0; i < s.size(); ++i) {
        if(isalpha(s[i])) {
            if(tmp.size()) {
                u[++n] = ID(tmp);
                tmp.clear();
            }
        }
        tmp += s[i];
    }
}

int two[2005] = {1};

string s, t;

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int T; cin >> T;
    int kase = 0;
    for(int i = 1; i <= 2000; ++i) two[i] = two[i - 1] * 2 % MOD;
    while(T--) {
        cin >> s >> t;
        cnt = 0; mp.clear();
        handle(n, s, u);
        handle(m, t, v);

        memset(dp, 0, sizeof dp);
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                if(u[i] == v[j])
                    add(dp[i][j], 2 * dp[i - 1][j - 1] + 2);
                else
                    add(dp[i][j], dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1]);
            }
        }
        int ans = ((two[n] + two[m] - 2 - dp[n][m]) % MOD + MOD) % MOD;
        printf("Case %d: %d\n", ++kase, ans);
    }
    return 0;
}

---

$O(nlogn)$

$这里就用到LCS和LIS的转化关系了$
$考虑a[i],b[j]相同的连一条边, 如果是公共子序列的话, 我们发现边不会交叉, 我们可以很惊喜的发现这就是LIS了, 问题由二维变成了一维$
$3.\ 考虑dp[i]:=以a[i]结尾上升子序列个数$
$BIT优化下转移过程就好了$
$ans_{相同的子序列个数} = \sum_{i=1}^n dp[i]$
$复杂度为O(nlogn)$

代码:

//
//  Created by TaoSama on 2015-11-22
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 10000007;

int n, m, c, a[N];
int b[N], dp[N];

int cnt = 0;
map<string, int> mp;

int ID(string &s) {
    if(mp.count(s)) return mp[s];
    mp[s] = ++cnt;
    return -1;
}

void add(int &x, int y) {
    x += y;
    if(x < 0) x += MOD;
    else if(x >= MOD) x -= MOD;
}

void update(int i, int v) {
    for(; i <= cnt; i += i & -i) add(b[i], v);
}

int sum(int i) {
    int ret = 0;
    for(; i; i -= i & -i) add(ret, b[i]);
    return ret;
}

void handle(int &n, string &s, bool flag) {
    n = c = 0; s += 'A';
    string tmp; tmp += s[0];
    for(int i = 1; i < s.size(); ++i) {
        if(isalpha(s[i])) {
            ++n;
            int id = ID(tmp);
            if(flag && ~id) a[++c] = id;
            tmp.clear();
        }
        tmp += s[i];
    }
}

int two[2005] = {1};

string s, t;

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
//  ios_base::sync_with_stdio(0);

    int T; cin >> T;
    int kase = 0;
    for(int i = 1; i <= 2000; ++i) two[i] = two[i - 1] * 2 % MOD;
    while(T--) {
        cin >> s >> t;
        cnt = 0; mp.clear();
        handle(n, s, false);
        handle(m, t, true);

        memset(dp, 0, sizeof dp);
        memset(b, 0, sizeof b);
        for(int i = 1; i <= c; ++i) {
            dp[i] = sum(a[i]) + 1;
            update(a[i], dp[i]);
        }
        int ans = ((two[n] + two[m] - 2 - 2 * sum(cnt)) % MOD + MOD) % MOD;
        printf("Case %d: %d\n", ++kase, ans);
    }
    return 0;
}