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：
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <stack> 16 #include <set> 17 #include <bitset> 18 #include <functional> 19 #include <numeric> 20 #include <stdexcept> 21 #include <utility> 22 23 using namespace std; 24 25 #define mem0(a) memset(a, 0, sizeof(a)) 26 #define mem_1(a) memset(a, -1, sizeof(a)) 27 #define lson l, m, rt << 1 28 #define rson m + 1, r, rt << 1 | 1 29 #define define_m int m = (l + r) >> 1 30 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 31 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 32 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 33 #define rep_down1(a, b) for (int a = b; a > 0; a--) 34 #define all(a) (a).begin(), (a).end() 35 #define lowbit(x) ((x) & (-(x))) 36 #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) {} 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 39 #define pchr(a) putchar(a) 40 #define pstr(a) printf("%s", a) 41 #define sstr(a) scanf("%s", a) 42 #define sint(a) scanf("%d", &a) 43 #define sint2(a, b) scanf("%d%d", &a, &b) 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 45 #define pint(a) printf("%d ", a) 46 #define test_print1(a) cout << "var1 = " << a << endl 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 49 #define mp(a, b) make_pair(a, b) 50 #define pb(a) push_back(a) 51 52 typedef long long LL; 53 typedef pair<int, int> pii; 54 typedef vector<int> vi; 55 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 58 const int maxn = 1e3 + 7; 59 const int md = 1e9 + 7; 60 const int inf = 1e9 + 7; 61 const LL inf_L = 1e18 + 7; 62 const double pi = acos(-1.0); 63 const double eps = 1e-6; 64 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 68 template<class T>T condition(bool f, T a, T b){return f?a:b;} 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 70 int make_id(int x, int y, int n) { return x * n + y; } 71 72 struct Graph { 73 vector<vector<int> > G; 74 void clear() { G.clear(); } 75 void resize(int n) { G.resize(n + 2); } 76 void add(int u, int v) { G[u].push_back(v); } 77 vector<int> & operator [] (int u) { return G[u]; } 78 }; 79 Graph G; 80 81 int mp[12][12 * maxn]; 82 int dp[maxn], a[12][maxn], b[12 * maxn]; 83 bool vis[maxn]; 84 85 int dfs(int pos) { 86 if (vis[pos]) return dp[pos]; 87 vis[pos] = true; 88 int sz = G[pos].size(); 89 rep_up0(i, sz) { 90 max_update(dp[pos], dfs(G[pos][i]) + 1); 91 } 92 return dp[pos]; 93 } 94 95 int main() { 96 //freopen("in.txt", "r", stdin); 97 int n, m; 98 while (cin >> n >> m) { 99 rep_up0(i, n) { 100 rep_up0(j, m) { 101 sint(a[i][j]); 102 b[i * m + j] = a[i][j]; 103 } 104 } 105 sort(b, b + n * m); 106 int sz = unique(b, b + n * m) - b; 107 mem0(mp); 108 rep_up0(i, n) { 109 rep_up0(j, m) { 110 a[i][j] = lower_bound(b, b + sz, a[i][j]) - b; 111 mp[i][a[i][j]] = j + 1; 112 } 113 } 114 G.clear(); 115 G.resize(m); 116 rep_up0(i, m) dp[i] = 1; 117 rep_up0(i, m) { 118 rep_up1(j, n - 1) { 119 if (!mp[j][a[0][i]]) { 120 dp[i] = -inf; 121 break; 122 } 123 } 124 } 125 rep_up0(i, m) { 126 for (int j = i + 1; j < m; j ++) { 127 bool ok = true; 128 int x = a[0][i], y = a[0][j]; 129 rep_up1(k, n - 1) { 130 if (!(mp[k][x] && mp[k][y] && mp[k][x] < mp[k][y])) { 131 ok = false; 132 break; 133 } 134 } 135 if (ok) G[i].push_back(j); 136 } 137 } 138 int ans = 0; 139 mem0(vis); 140 rep_up0(i, m) max_update(ans, dfs(i)); 141 cout << ans << endl; 142 } 143 return 0; 144 }