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A.Mr. Kitayuta, the Treasure Hunter
很显然的一个DP，30000的数据导致使用map+set会超时。题解给了一个非常实用的做法，由于每个点有不超过250种状态，并且这些状态都是以包含d连续的一段数字，那么可以以对d的偏移量作为状态。这算是很常见的一个优化了。
#include<bits/stdc++.h> using namespace std; const int INF = 30009; int f[INF][500],a[INF]; int n, d, ans = 1, x; int main() { scanf ("%d %d", &n, &d); for (int i = 1; i <= n; i++) { scanf ("%d", &x); a[x]++; } f[d][250] = a[d] + 1; for (int i = d; i <= x; i++) for (int j = 0; j < 500; j++) { if (f[i][j] == 0) continue; int k = j - 250 + d, val = f[i][j]; ans = max (ans, val); for (int t = max (1, k - 1); t <= k + 1; t++) if (i + t <= x) f[i + t][t - d + 250] = max (f[i + t][t - d + 250], val + a[i + t]); } printf ("%d\n", ans - 1); }
[Problem] Given an integer n and m pairs of integers (ai, bi) (1 ≤ ai, bi ≤ n), find the minimum number of edges in a directed graph that satisfies the following condition:
对于一个n个点的强连通图最少需要n条边，而非强连通的n个有边连接的点最少需要n-1条边。那么只要判断连通起来的k个点组成的块是不是有环，如果有环需要k条边，无环需要k-1条边。在连完一个连通的块后将其合并看成一个点。直到所有点都遍历过。
#include<bits/stdc++.h> using namespace std; const int INF = 120009; struct Edge { int v, next; } edge[INF]; int head[INF], ncnt ; int f[INF], vis[INF], cycle[INF], siz[INF]; int n, m, ans; inline void adde (int u, int v) { edge[++ncnt].v = v, edge[ncnt].next = head[u]; head[u] = ncnt; } inline int find (int x) { if (f[x] == x) return x; return f[x] = find (f[x]); } inline void uin (int x, int y) { siz[find (x)] += siz[find (y)]; f[find (y)] = find (x); } inline bool dfs (int u) { vis[u] = 1; for (int i = head[u]; i; i = edge[i].next) { int v = edge[i].v; if (!vis[v]) dfs (v); if (vis[v] == 1) cycle[find (u)] = 1; } vis[u] = 2; } int main() { scanf ("%d %d", &n, &m); for (int i = 1; i <= n; i++) f[i] = i, siz[i] = 1; for (int i = 1, x, y; i <= m; i++) { scanf ("%d %d", &x, &y); adde (x, y); if (find (x) != find (y) ) uin (x, y); } for (int i = 1; i <= n; i++) if (!vis[i]) dfs (i); for (int i = 1; i <= n; i++) { if (find (i) == i) ans += siz[i] - 1; ans += cycle[i]; } printf ("%d\n", ans); }
D.Mr. Kitayuta‘s Colorful Graph
题意：一个n个点m条边的无向图中所有的边都有一个颜色，有q个询问，对每个询问（u，v）输出从u到v的纯色路径条数。(n,m,q<1e5)
首先应该先把所有相同颜色边连接的块做一次并查集，然后对所有询问（a，b）对a，b中有着较少的度的点的颜色查找b是否与a在一个集合。
unordered_map能够在让查找b的时间降到O（1）
#include<bits/stdc++.h> using namespace std; const int INF = 100009; unordered_map<int, int> f[INF], old[INF]; int n, m, q; inline int find (int x, int c) { if (f[x][c] < 0) return x; return f[x][c] = find (f[x][c], c); } inline void merge (int x, int y, int c) { if (f[x].find (c) == f[x].end() ) f[x][c] = -1; if (f[y].find (c) == f[y].end() ) f[y][c] = -1; x = find (x, c), y = find (y, c); if (x == y) return; f[y][c] += f[x][c]; f[x][c] = y; } int main() { scanf ("%d %d", &n, &m); for (int i = 1, x, y, c; i <= m; i++) { scanf ("%d %d %d", &x, &y, &c); merge (x, y, c); } scanf ("%d", &q); int a, b; while (q--) { scanf ("%d %d", &a, &b); if (f[a].size() > f[b].size() ) swap (a, b); if (old[a].find (b) == old[a].end() ) { int ans = 0; for (auto &c : f[a]) if (f[b].find (c.first) != f[b].end() && find (a, c.first) == find (b, c.first) ) ans++; old[a][b] = ans; } printf ("%d\n", old[a][b]); } }
Codeforces Round #286 (Div. 1) 解题报告
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原文地址：http://www.cnblogs.com/keam37/p/4237581.html
