最长公共子序列【线段树版的LCS】

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最新推荐文章于 2022-09-22 10:38:34 发布

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分类专栏：知识点整理模板集（陆续更新）文章标签： insert build vector struct

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#include<iostream> #include<cstring> #include<algorithm> #include<vector> #define N 200008 #define M using namespace std; int hash1[N], hash2[N], x[N], y[N], dp[N], next1[N], next2[N]; vector <int> v[N]; struct LTree { int a, b, big; }T[N]; int e1(int a) { int i = a; while (next1[i] != i) i = e1(next1[i]); return next1[a] = i; } int e2(int a) { int i = a; while (next2[i] != i) i = e2(next2[i]); return next2[a] = i; } void f1(int a, int b) { int i = e1(b); hash1[i] = a; next1[i] = i + 1; return; } void f2(int a, int b) { int i = e2(b); hash2[i] = a; next2[i] = i + 1; return; } void build(int i) { T[i].big = 0; if (T[i].a != T[i].b) { int t = i*2+1; T[t].a = T[i].a; T[t].b = (T[i].a+T[i].b)/2; build(t); t++; T[t].a = (T[i].a+T[i].b)/2 + 1; T[t].b = T[i].b; build(t); } } void insert(int i, int x, int k) { if (x < T[i].a || x > T[i].b) return; T[i].big = max(T[i].big, k); if (T[i].a == T[i].b) return; if (x <= (T[i].a+T[i].b)/2) insert(i*2+1, x, k); else insert(i*2+2, x, k); } int ask(int i, int x) { if (x < T[i].a) return 0; if (T[i].b <= x) return T[i].big; if (x <= (T[i].a+T[i].b)/2) return ask(i*2+1, x); else return max(T[i*2+1].big, ask(i*2+2, x)); } int main() { int t, test, n, m; int i, j; int a, b, all, count, temp, ad, ans, big; scanf("%d", &test); for (t=1; t<=test; t++) { scanf("%d%d", &n, &m); memset(hash1, 0, sizeof(hash1)); memset(hash2, 0, sizeof(hash2)); all = n*m; for (i=0; i<N; i++) { next1[i] = i; next2[i] = i; } for (i=0; i<all; i++) { scanf("%d%d", &a, &b); f1(a, b); } for (i=0; i<all; i++) { scanf("%d%d", &a, &b); f2(a, b); } for (i=0, count=1; i<N && count<=all; i++) if (hash1[i]) x[count++] = hash1[i]; for (i=0, count=1; i<N && count<=all; i++) if (hash2[i]) y[count++] = hash2[i]; for (i=1; i<=all; i++) v[i].clear(); for (i=1; i<=all; i++) v[y[i]].push_back(i); T[0].a = 1; T[0].b = all; build(0); ans = 0; for (i=1; i<=all; i++) { temp = (int)v[x[i]].size(); for (j=temp-1; j>=0; j--) { ad = v[x[i]][j]; big = ask(0, ad-1) + 1; insert(0, ad, big); ans = max(ans, big); } } printf("Case #%d: %d/n", t, ans); } return 0; }