有一个N×M的单位方格中，其中有些方格是水塘，其他方格是陆地。如果要用1×2的矩阵区覆盖（覆盖过程不容许有任何部分重叠）这个陆地，那么最多可以覆盖多少陆地面积。

输入文件的第一行是两个整数N，M  (1<=N，M<=100)，第二行为一个整数K( K<=50)，接下来的K行，每行两个整数X，Y表示K个水塘的行列位置。(1<=X<=N，1<=Y<=M)。

输出所覆盖的最大面积块（1×2面积算一块）。

4 4
6
1 1
1 4
2 2
4 1
4 2
4 4

4

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
#define rep(i, s, t) for(int i = (s); i <= (t); i++)
#define dwn(i, s, t) for(int i = (s); i >= (t); i--)int read() {int x = 0, f = 1; char c = getchar();while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }return x * f;
}#define maxn 20010
#define maxm 320010
#define oo 2147483647struct Edge {int from, to, flow;Edge() {}Edge(int _1, int _2, int _3): from(_1), to(_2), flow(_3) {}
};
struct Dinic {int n, m, s, t, head[maxn], nxt[maxm];Edge es[maxm];int Q[maxn], hd, tl, vis[maxn];int cur[maxn];void init() {m = 0; memset(head, -1, sizeof(head));return ;}void setn(int _) {n = _;return ;}void AddEdge(int a, int b, int c) {es[m] = Edge(a, b, c); nxt[m] = head[a]; head[a] = m++;es[m] = Edge(b, a, 0); nxt[m] = head[b]; head[b] = m++;return ;}bool BFS() {memset(vis, 0, sizeof(vis));hd = tl = 0; Q[++tl] = s; vis[s] = 1;while(hd < tl) {int u = Q[++hd];for(int i = head[u]; i != -1; i = nxt[i]) {Edge& e = es[i];if(!vis[e.to] && e.flow) {vis[e.to] = vis[u] + 1;Q[++tl] = e.to;}}}return vis[t] > 1;}int DFS(int u, int a) {if(u == t || !a) return a;int flow = 0, f;for(int &i = cur[u]; i != -1; i = nxt[i]) {Edge& e = es[i];if(vis[e.to] == vis[u] + 1 && (f = DFS(e.to, min(a, e.flow)))) {flow += f; a -= f;e.flow -= f; es[i^1].flow += f;if(!a) return flow;}}return flow;}int MaxFlow(int _s, int _t) {s = _s; t = _t;int flow = 0;while(BFS()) {rep(i, 1, n) cur[i] = head[i];flow += DFS(s, oo);}return flow;}
} sol;#define maxl 65
int n, m, K, CntP;
bool Map[maxl][maxl];
struct Node {int id;Node(): id(0) {}int p() { return id ? id : id = ++CntP; }
} nsi[maxl][maxl], nso[maxl][maxl], SS, TT;int main() {n = read(); m = read(); K = read();rep(i, 1, K) {int x = read(), y = read();Map[x][y] = 1;}sol.init();rep(i, 1, n) rep(j, 1, m) if(!Map[i][j]){if(i < n && !Map[i+1][j]) sol.AddEdge(nso[i][j].p(), nsi[i+1][j].p(), 1), sol.AddEdge(nso[i+1][j].p(), nsi[i][j].p(), 1);if(j < n && !Map[i][j+1]) sol.AddEdge(nso[i][j].p(), nsi[i][j+1].p(), 1), sol.AddEdge(nso[i][j+1].p(), nsi[i][j].p(), 1);if((i ^ j) & 1) sol.AddEdge(SS.p(), nsi[i][j].p(), 1);else sol.AddEdge(nso[i][j].p(), TT.p(), 1);sol.AddEdge(nsi[i][j].p(), nso[i][j].p(), 1);}sol.setn(CntP);printf("%d\n", sol.MaxFlow(SS.p(), TT.p()));return 0;
}