[csu/coj 1078]多个序列的最长公共子序列

题意：给n个序列，同一个序列里面元素互不相同，求它们的最长公共子序列。

思路：任取一个序列，对于这个序列里面的两个数ai,aj(i<j)，如果对于其它每一个序列，都出现过ai,aj，且ai在aj之前出现，那么i到j连一条长度为1的有向边，完美转化为DAG最长路。需要注意：对于某个数，如果某个序列没出现那么这个点的答案应该为-INF，表示这个点表示的状态不合法。

代码：

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <stack>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e3 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 struct Graph {
     vector<vector<int> > G;
     void clear() { G.clear(); }
     void resize(int n) { G.resize(n + ); }
     void add(int u, int v) { G[u].push_back(v); }
     vector<int> & operator [] (int u) { return G[u]; }
 };
 Graph G;

 int mp[][ * maxn];
 int dp[maxn], a[][maxn], b[ * maxn];
 bool vis[maxn];

 int dfs(int pos) {
     if (vis[pos]) return dp[pos];
     vis[pos] = true;
     int sz = G[pos].size();
     rep_up0(i, sz) {
         max_update(dp[pos], dfs(G[pos][i]) + );
     }
     return dp[pos];
 }

 int main() {
     //freopen("in.txt", "r", stdin);
     int n, m;
     while (cin >> n >> m) {
         rep_up0(i, n) {
             rep_up0(j, m) {
                 sint(a[i][j]);
                 b[i * m + j] = a[i][j];
             }
         }
         sort(b, b + n * m);
         int sz = unique(b, b + n * m) - b;
         mem0(mp);
         rep_up0(i, n) {
             rep_up0(j, m) {
                 a[i][j] = lower_bound(b, b + sz, a[i][j]) - b;
                 mp[i][a[i][j]] = j + ;
             }
         }
         G.clear();
         G.resize(m);
         rep_up0(i, m) dp[i] = ;
         rep_up0(i, m) {
             rep_up1(j, n - ) {
                 if (!mp[j][a[][i]]) {
                     dp[i] = -inf;
                     break;
                 }
             }
         }
         rep_up0(i, m) {
             for (int j = i + ; j < m; j ++) {
                 bool ok = true;
                 int x = a[][i], y = a[][j];
                 rep_up1(k, n - ) {
                     if (!(mp[k][x] && mp[k][y] && mp[k][x] < mp[k][y])) {
                         ok = false;
                         break;
                     }
                 }
                 if (ok) G[i].push_back(j);
             }
         }
         int ans = ;
         mem0(vis);
         rep_up0(i, m) max_update(ans, dfs(i));
         cout << ans << endl;
     }
     return ;
 }