解题报告
http://blog.csdn.net/juncoder/article/details/38172083
题意：
M×N的矩阵，k个点被标记，用2×1的木板最多能够放置多少个。
思路：
把标记的格子除外，链接相邻的两个格子，然后最大匹配出来的是二分图的两倍。
c++TLE了，G++1700+过了，理论上匈牙利算法的时间复杂度是n^3。就应该超时。可能数据弱吧。
另一种建图方式就是建成二分图，将矩阵中的点奇偶分。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,m,k,mmap[2050][2050],edge[2050][2050],pre[2050],vis[2050],kk;
int dx[]= {1,-1,0,0};
int dy[]= {0,0,1,-1};
int dfs(int x)
{
for(int i=1; i<=kk; i++) {
if(!vis[i]&&edge[x][i]) {
vis[i]=1;
if(pre[i]==-1||dfs(pre[i])) {
pre[i]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,a,b;
while(~scanf("%d%d%d",&m,&n,&k)) {
memset(mmap,0,sizeof(mmap));
memset(pre,-1,sizeof(pre));
memset(edge,0,sizeof(edge));
for(i=1; i<=k; i++) {
scanf("%d%d",&b,&a);
mmap[a][b]=-1;
}
if(n*m%2!=k%2) {
printf("NO
");
continue;
}
kk=0;
for(i=1; i<=m; i++) {
for(j=1; j<=n; j++) {
if(!mmap[i][j])
mmap[i][j]=++kk;
}
}
if(kk%2!=0) {
printf("NO
");
continue;
}
int l=0;
for(i=1; i<=m; i++) {
for(j=1; j<=n; j++) {
if(mmap[i][j]!=-1)
for(l=0; l<4; l++) {
int x=i+dx[l];
int y=j+dy[l];
if(x>=1&&x<=m&&y>=1&&y<=n&&mmap[x][y]!=-1) {
edge[mmap[i][j]][mmap[x][y]]=1;
}
}
}
}
int ans=0;
for(i=1; i<=kk; i++) {
memset(vis,0,sizeof(vis));
ans+=dfs(i);
}
if(ans==kk)
printf("YES
");
else printf("NO
");
}
return 0;
}
Chessboard
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 13140 | Accepted: 4105 |
Description
Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of cards with size 1 * 2 to cover the board. However, she thinks it too easy to bob, so she makes some holes on the board (as shown in the
figure below).
We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below:
1. Any normal grid should be covered with exactly one card.
2. One card should cover exactly 2 normal adjacent grids.
Some examples are given in the figures below:
A VALID solution.
An invalid solution, because the hole of red color is covered with a card.
An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.
We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below:
1. Any normal grid should be covered with exactly one card.
2. One card should cover exactly 2 normal adjacent grids.
Some examples are given in the figures below:
A VALID solution.
An invalid solution, because the hole of red color is covered with a card.
An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.
Input
There are 3 integers in the first line: m, n, k (0 < m, n <= 32, 0 <= K < m * n), the number of rows, column and holes. In the next k lines, there is a pair of integers (x, y) in each line, which represents a hole in the y-th row, the x-th column.
Output
If the board can be covered, output "YES". Otherwise, output "NO".
Sample Input
4 3 2 2 1 3 3
Sample Output
YES
Hint
A possible solution for the sample input.
Source
POJ Monthly,charlescpp