赛时怎么忘记看了

这道题题意很明确,就是求最大独立集。对于 a=b=1a=b=1 的情况大佬们肯定都会(不会求的自行学习二分图),所以我们可以根据 aabb 的奇偶性按照某种策略给棋盘黑白染色,满足能够一步之内相互到达的点颜色不同就可以。

首先 aabb 一奇一偶的情况是最简单的,由于走完之后横纵坐标之和的奇偶性会改变,直接把棋盘黑白交替染色就可以。

第二个是 aabb 都是奇数的情况,这个时候可以按照行或者列的奇偶性染色,因为横坐标和纵坐标的奇偶性会改变。

最后一个是 aabb 都是偶数的情况。我们想一想怎么把情况转化成上面两种,不难想到可以把 aabb 每次都同时除掉 22,但这个时候我们把棋盘变小了,横纵坐标也要同时除 22 才能保证原来的位置关系,然后你会发现就做完了。

剩下的上一个 dinic 板子做完了。

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#include<bits/stdc++.h>
using namespace std;
#define int long long
struct flowGraph
{
struct edge {
int v, nxt, cap, flow;
} edges[2222222];
int head[2222222], cnt = 0;
int n, S, T;
int maxflow = 0;
int dist[2222222], cur[2222222];
void init(int _n, int _S, int _T) {
memset(head, -1, sizeof(head));
cnt = 0;
S = _S;
T = _T;
n = _n;
}
void addedge(int u, int v, int w) {
edges[cnt] = {v, head[u], w, 0};
head[u] = cnt++;
edges[cnt] = {u, head[v], 0, 0};
head[v] = cnt++;
}
bool bfs() {
queue<int> q;
memset(dist, 0, sizeof(int) * (n + 1));
dist[S] = 1;
q.push(S);
while (q.size()) {
int u = q.front();
q.pop();
for (int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if ((!dist[v]) && (edges[i].cap > edges[i].flow)) {
dist[v] = dist[u] + 1;
q.push(v);
}
}
}
return dist[T];
}
int dfs(int u, int flow) {
if ((u == T) || (!flow)) {
return flow;
}
int res = 0;
for (int &i = cur[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v, f;
if ((dist[v] == dist[u] + 1) && (f = dfs(v, min(flow - res, edges[i].cap - edges[i].flow)))) {
res += f;
edges[i].flow += f;
edges[i ^ 1].flow -= f;
if (res == flow) {
return res;
}
}
}
return res;
}
int dinic() {
while (bfs()) {
memcpy(cur, head, sizeof(int) * (n + 1));
maxflow += dfs(S, 0x3f3f3f3f3f3f);
}
return maxflow;
}
} g;
int dir[4][2] = {{1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int n, a, b;
bool calc1(int x, int y)
{
int ta = a, tb = b;
while (ta % 2 == 0 && tb % 2 == 0) {
ta /= 2;
tb /= 2;
x /= 2;
y /= 2;
}
if ((ta + tb) % 2) {
return (x + y) % 2;
} else {
return x % 2;
}
}
int calc2(int x, int y)
{
return x * n + y;
}
signed main()
{
cin >> n >> a >> b;
vector<vector<char>> s(n + 1, vector<char>(n + 1));
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
cin >> s[i][j];
}
}
g.init(n * n + n + 2, 0, n * n + n + 1);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (s[i][j] == '#') {
continue;
}
if (!calc1(i, j)) {
g.addedge(0, calc2(i, j), 1);
for (int k = 0; k < 4; k++) {
int dx = i + dir[k][0] * a;
int dy = j + dir[k][1] * b;
if (1 <= dx && dx <= n && 1 <= dy && dy <= n && s[dx][dy] != '#') {
g.addedge(calc2(i, j), calc2(dx, dy), 1);
}
dx = i + dir[k][0] * b;
dy = j + dir[k][1] * a;
if (1 <= dx && dx <= n && 1 <= dy && dy <= n && s[dx][dy] != '#') {
g.addedge(calc2(i, j), calc2(dx, dy), 1);
}
}
} else {
g.addedge(calc2(i, j), n * n + n + 1, 1);
}
}
}
g.dinic();
vector<bool> vis(n * n + n + 1, 0);
queue<int> q;
q.push(0);
vis[0] = 1;
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = g.head[u]; ~i; i = g.edges[i].nxt) {
if (g.edges[i].cap > g.edges[i].flow && !vis[g.edges[i].v]) {
vis[g.edges[i].v] = 1;
q.push(g.edges[i].v);
}
}
}
vector<string> ans(n + 1, string(n + 1, '.'));
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (s[i][j] == '#') {
ans[i][j] = '#';
continue;
}
if (!calc1(i, j)) {
if (vis[calc2(i, j)]) {
ans[i][j] = 'o';
} else {
ans[i][j] = '.';
}
} else {
if (!vis[calc2(i, j)]) {
ans[i][j] = 'o';
} else {
ans[i][j] = '.';
}
}
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
cout << ans[i][j];
}
cout << '\n';
}
return 0;
}