- 題目鏈接:http://poj.org/problem?id=2446
- 題意:給你一個 m*n的棋盤,要你用1×2大小的卡片覆蓋它,並且每個方格都只能被一個卡片覆蓋。其中k個方格不能放卡片。(0 < m, n <= 32, 0 <= K < m * n)
- 思路:將這個棋盤想象成國際象棋的黑白相間的棋盤,黑格和其上下左右的四個白格匹配即可建成一個二分圖,當然不能放卡片的那幾個格子要過濾掉。
- 注意:這裏是 m行 n列!先輸入的是m。在程序裏將m和n調換也可以。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <math.h>
#define pi acos(-1)
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
const int INF = 0x3f3f3f3f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 32 + 10;
const int maxm = 32 + 10;
const int maxv = 10000 + 10;
const int maxe = 20000 + 10;
const int mod = 1e9 + 7;
int mp[40][40];
int dir[4][2] = {0,1, 0,-1, 1,0, -1,0};
int S, T;
int head[maxe], tot=0, nowedge[maxe];
int dis[maxv];
int vis[maxv];
struct Edge
{
int to, next, cap;
}es[maxe];
int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
void init()
{
memset(head, -1, sizeof(head));
memset(vis, 0, sizeof(vis));
tot=0;
}
void add(int u, int v, int vol)
{
es[tot].to = v;
es[tot].cap = vol;
es[tot].next = head[u];
head[u] = tot++;
es[tot].to = u;
es[tot].cap = 0;
es[tot].next = head[v];
head[v] = tot++;
}
bool BFS()
{
queue<int> que;
memset(dis, -1, sizeof(dis));
dis[S] = 0;
que.push(S);
int cur, v;
while(!que.empty())
{
cur = que.front(); que.pop();
for(int i=head[cur]; i!=-1; i=es[i].next){
v = es[i].to;
if(dis[v]==-1 && es[i].cap>0){
dis[v] = dis[cur]+1;
que.push(v);
}
}
}
if(dis[T] == -1) return false;
else return true;
}
LL DFS(int cur, LL low)
{
LL res=0, add=0;
if(cur == T) return low;
for(int i=nowedge[cur]; i!=-1; i=es[i].next){
nowedge[cur] = i;
if(dis[es[i].to]==dis[cur]+1
&& es[i].cap>0
&& (add = DFS(es[i].to, min(low, (LL)es[i].cap))) ){
es[i].cap -= add;
es[i^1].cap += add;
low -= add;
res += add;
if(low == 0) break;
}
}
if(res == 0) dis[cur] = 0;
return res;
}
LL MAXFLOW()
{
LL ans=0 , tmp=0;
while(BFS()){
for(int i=S; i<=T; i++){
nowedge[i] = head[i];
}
if(tmp = DFS(S, INF))
ans += tmp;
}
return ans;
}
int left_net(int cur)
{
vis[cur] = 1;
int ans = 1;
for(int i=head[cur]; i!=-1; i = es[i].next){
if(es[i].cap > 0 && !vis[es[i].to]) ans += left_net(es[i].to);
}
return ans;
}
int main()
{
int n, m, k;
init();
scanf("%d%d%d", &n, &m, &k);
S = 0;
T = n*m+1;
int tmp1, tmp2;
for(int i=0; i<k; i++){
scanf("%d%d", &tmp2, &tmp1);
mp[tmp1][tmp2] = 1;
}
for(int i=1; i<=n; i++){
for(int j=1; j<=m; j++){
if(mp[i][j]) continue;
if((i+j)%2==1){
add(S, (i-1)*m+j, 1);
for(int dd=0; dd<4; dd++){
int x = i+dir[dd][0];
int y = j+dir[dd][1];
if(mp[x][y] != 1 && x>=1 && x<= n && y>=1 && y<=m){
add((i-1)*m+j, (x-1)*m+y, INF);
}
}
}
else add((i-1)*m+j, T, 1);
}
}
LL ans = MAXFLOW();
if(ans*2 != (n*m-k)) printf("NO\n");
else printf("YES\n");
}