2038. Minimum Vertex Cover
Time limit: 1.0 second
Memory limit: 64 MB
Memory limit: 64 MB
A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex of the set. A minimum vertex cover is a vertex cover with minimal cardinality.
Consider a set of all minimum vertex covers of a given bipartite graph. Your task is to divide all vertices of the graph into three sets. A vertex is in setN (“Never”) if there is no minimum vertex cover containing this vertex. A vertex is in set A (“Always”) if it is a part of every minimum vertex cover of the given graph. If a vertex belongs neither to N nor to A, it goes to the set E (“Exists”).
Input
The first line of input contains three integers n, m, k: the size of the first vertex set of the bipartite graph, the size of the second vertex set and the number of edges (1 ≤ n, m ≤ 1000; 0 ≤ k ≤ 106). Next k lines contain pairs of numbers of vertices, connected by an edge. First number denotes a vertex from the first set, second — from the second set. Vertices in each set are numbered starting from one. No pair of vertices is connected by more than one edge.
Output
On the first line, print a sequence of n letters ‘N’, ‘E’, ‘A’ without spaces. The letter on position i corresponds to the set containing i-th vertex of the first set. The second line must contain the answer for the second vertex set in the same format.
Sample
| input | output |
|---|---|
11 9 22 1 1 1 2 1 3 1 8 1 9 2 1 2 3 3 2 3 4 4 3 4 5 5 2 5 4 5 6 6 6 6 7 7 5 7 7 8 7 9 7 10 7 11 7 |
AEEEEEENNNN EEEEEEANN |
题解：
看了网上 http://blog.csdn.net/u011699990/article/details/45257071 的题解才会的
关键就是确定这个点是否在最小点覆盖的条件是 ： 如果这个点是一个已盖点，从他相连的所有点中只要存在一个是未盖点，那么这个点一定是在最小点覆盖上的（这个结论很神奇，也很容易想通）
那么一旦确定了某个点是最小点覆盖上的点之后，他所对应的另外一个匹配点就一定不会在最小点覆盖了，如此循环，我们就能最终确定哪些点是一定在最小点覆盖上了，可以看看代码理解，代码写了注释
#include <bits/stdc++.h>
#define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a))
#define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a))
#define pb push_back
#define mp make_pair
#define sf scanf
#define pf printf
#define two(x) (1<<(x))
#define clr(x,y) memset((x),(y),sizeof((x)))
#define dbg(x) cout << #x << "=" << x << endl;
#define lowbit(x) ((x)&(-x))
const int mod = 1e9 + 7;
int mul(int x,int y){return 1LL*x*y%mod;}
int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;}
inline 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;}
using namespace std;
const int maxn = 1e3 + 15;
vector < int > LE[maxn] , RE[maxn] ;
int linkL[maxn],linkR[maxn],N,M,K,ansL[maxn],ansR[maxn],vis[maxn];
int DFS( int x ){
for( auto v : LE[x] ){
if( vis[v] ) continue;
vis[v] = 1;
if( linkR[v] == -1 || DFS( linkR[v] ) ){
linkL[x] = v , linkR[v] = x;
return 1;
}
}
return 0;
}
int main( int argc , char * argv[] ){
N=read(),M=read(),K=read();
rep(i,1,K){
int u = read() , v = read();
LE[u].pb( v ); RE[v].pb( u );
}
clr( linkL , -1 ) ; clr( linkR , -1 );
// 进行二分图最大匹配
rep(i,1,N){
rep(j,1,M) vis[j]=0;
DFS( i );
}
// 初始认为所有匹配的点都是 E , 对于未匹配点，加入队列中
queue < int > Q;
rep(i,1,N) if(~linkL[i]) ansL[i] = 1; else Q.push( i );
rep(i,1,M) if(~linkR[i]) ansR[i] = 1; else Q.push( i + N );
// 对于一定在最小点覆盖的点满足的条件即 ： 这个匹配点连接的所有边中存在非匹配点 , 即，如果存在的话，就必须选这个点
// 开始跑
// 注意队列中的点都是未匹配点
while(!Q.empty()){
int x = Q.front() ; Q.pop();
// 是左边的点
if( x <= N ){
for( auto v : LE[x] ){
// 右边的点是一个匹配点同时没有check过
if( ~linkR[v] && ansR[v] != 2 ){
ansR[v] = 2; // 这个点必选
ansL[linkR[v]] = 0; // 那么他所对应的左边的匹配点一定不选
Q.push( linkR[v] ); // 左边那个点不选了，加入到队列中
}
}
}else{
for( auto v : RE[x - N] ){
// 左边的点是一个匹配点同时没有check过
if( ~linkL[v] && ansL[v] != 2 ){
ansL[v] = 2; // 这个点必选
ansR[linkL[v]] = 0; // 那么他所对应的右边的匹配点一定不选
Q.push( linkL[v] + N ); // 左边那个点不选了，加入到队列中
}
}
}
}
rep(i,1,N) if(ansL[i] == 0) putchar('N') ; else if( ansL[i] == 1 ) putchar('E') ; else putchar('A');
puts("");
rep(i,1,M) if(ansR[i] == 0) putchar('N') ; else if( ansR[i] == 1 ) putchar('E') ; else putchar('A');
puts("");
return 0;
}