[網路流24題]P3355 騎士共存
阿新 • • 發佈:2021-10-22
二分圖/最大流,用計算機系統的記憶體訪問卡常
https://www.luogu.com.cn/problem/P3355
思路
程式碼
來看一看下面兩個存圖時的訪問順序
int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
和
int dx[8] = {1, 2, 2, 1, -1, -2, -2, -1};
int dy[8] = {-2, -1, 1, 2, -2, -1, 1, 2};
上面的寫法本地執行4000+,而下面的只需要 700+,在計系中學習過連續記憶體訪問可以提高程式執行效率,這可以快樂地從TLE變成AC!表示學到了,下次可以 next_permutation 一組最快的提交2333
二分圖卡常程式碼
#include <bits/stdc++.h> using namespace std; bool vis[40010]; int link[40010]; int n, m, ans; vector<int> mp[40010]; bool f[205][205]; /* int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};*/ const int dx[8] = {1, 2, 2, 1, -1, -2, -2, -1}; const int dy[8] = {-2, -1, 1, 2, -2, -1, 1, 2}; int dfs(int x) { for (int i = 0; i < mp[x].size(); i++) { int v = mp[x][i]; if (vis[v] == 0) { vis[v] = 1; if (link[v] == 0 || dfs(link[v]) == 1) { link[v] = x; return 1; } } } return 0; } int id(int x, int y) { return (x - 1) * n + y; } clock_t st, ed; // int node[N], tot; int main() { st = clock(); freopen("C:\\Users\\lenovo\\Downloads\\P3355_3.in", "r", stdin); cin >> n >> m; for (int i = 0; i < m; i++) { int x, y; cin >> x >> y; f[x][y] = 1; } // cout << "aaa"; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if ((i + j) % 2 || f[i][j]) continue; for (int k = 0; k < 8; k++) { int nx = i + dx[k], ny = j + dy[k]; if (nx > 0 && ny > 0 && nx <= n && ny <= n && !f[nx][ny]) mp[id(i, j)].push_back(id(nx, ny)); } } } for (int i = 1; i <= n * n; i++) { memset(vis, 0, sizeof(vis)); if (dfs(i)) ans++; } cout << n * n - m - ans << endl; ed = clock(); cout << (ed - st) << endl; }
正經Dinic
點選檢視程式碼
#include <bits/stdc++.h> #define endl '\n' #define IOS \ ios::sync_with_stdio(0); \ cin.tie(0); \ cout.tie(0) #define P pair<int, int> #define endl '\n' using namespace std; typedef long long ll; const int maxn = 200 * 200 + 10; const int N = 200 * 200 + 10; const int M = N << 5; const int INF = 0x3f3f3f3f; const ll inf = 1e18; ll dis[maxn]; bool vis[maxn]; struct node { int v, w, to; } edge[M * 2]; int pre[N], cnt, dep[N]; int S, T, head[N], sum; int n, m, q[N], cur[N]; void add(int u, int v, int w) { // cout << u << " " << v << endl; edge[cnt] = {v, w, head[u]}; head[u] = cnt++; edge[cnt] = {u, 0, head[v]}; head[v] = cnt++; } bool bfs() { for (int i = 0; i <= T; i++) dep[i] = 0; dep[S] = 1; int l = 0, r = 1; q[r] = S; while (l < r) { int u = q[++l]; for (int i = head[u]; i != -1; i = edge[i].to) { int v = edge[i].v; if (!dep[v] && edge[i].w) dep[v] = dep[u] + 1, q[++r] = v; } } return dep[T]; } int dfs(int u, int mi) { int res = 0; if (mi == 0 || u == T) return mi; for (int &i = cur[u]; i != -1; i = edge[i].to) { int v = edge[i].v; if (dep[u] + 1 == dep[v] && edge[i].w) { int minn = dfs(v, min(mi - res, edge[i].w)); edge[i].w -= minn; edge[i ^ 1].w += minn; res += minn; if (res == mi) return res; } } if (res == 0) dep[u] = 0; return res; } int dinic() { ll res = 0; while (bfs()) { memcpy(cur, head, sizeof(head)); // cout<<res<<endl; res += dfs(S, INF); } return res; } int id(int x, int y) { return (n * (x - 1) + y); } int mp[205][205], f; int dx[] = {-2, -1, 1, 2, 2, 1, -1, -2}; int dy[] = {1, 2, 2, 1, -1, -2, -2, -1}; int main() { freopen("C:\\Users\\lenovo\\Downloads\\P3355_3.in", "r", stdin); memset(head, -1, sizeof head); cin >> n >> m; int ans = n * n; S = ans + 1, T = S + 1; for (int i = 0, x, y; i < m; i++) { cin >> x >> y; mp[x][y] = 1; ans--; } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (mp[i][j]) continue; int now = id(i, j); if ((i+j) & 1) { if (!mp[i][j]) add(S, now, 1); } else { if (!mp[i][j]) add(now, T, 1); continue; } for (int k = 0; k < 8; k++) { int nx = i + dx[k], ny = j + dy[k]; if (nx < 1 || nx > n || ny < 1 || ny > n || mp[nx][ny]) continue; int to = id(nx, ny); add(now, to, INF); // printf("(%d,%d) :%d\n", i, j, id(i, j)); // printf("(%d,%d) :%d\n", nx, ny, id(nx, ny)); // add(now, id(nx, ny), INF); } } } int flow = dinic(); // cout << flow << endl; ans -= flow; cout << ans; return 0; }