[hungary][最小顶点覆盖][二分图]hdu 1054

听说此题很经典，有三种解法

这里三种解法讲的都很好

在此，写出二分图的最大匹配方法

Strategic Game

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5313    Accepted Submission(s): 2452

Problem Description

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob has to put for a given tree.

The input file contains several data sets in text format. Each data set represents a tree with the following description:

the number of nodes

the description of each node in the following format

node_identifier:(number_of_roads) node_identifier1 node_identifier2 … node_identifier

or

node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500). Every edge appears only once in the input data.

For example for the tree:

the solution is one soldier ( at the node 1).

The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following table:

Sample Input

4

0:(1) 1

1:(2) 2 3

2:(0)

3:(0)

5

3:(3) 1 4 2

1:(1) 0

2:(0)

0:(0)

4:(0)

Sample Output

1

2

Source

Southeastern Europe 2000

//最小顶点覆盖问题
//然而一直超时
//于是用vector
//然后就不超时了kao
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif
#define INF 0x3f3f3f3f
#define clock CLOCKS_PER_SEC
#define cle(x) memset(x,0,sizeof(x))
#define maxcle(x) memset(x,0x3f,sizeof(x))
#define mincle(x) memset(x,-1,sizeof(x))
#define maxx(x1,x2,x3) max(x1,max(x2,x3))
#define minn(x1,x2,x3) min(x1,min(x2,x3))
#define cop(a,x) memcpy(x,a,sizeof(a))
#define FROP "hdu"
#define C(a,b) next_permutation(a,b)
#define LL long long
#define smin(x,tmp) x=min(x,tmp)
#define smax(x,tmp) x=max(x,tmp)
using namespace std;
const int N = 1505;
vector<int>G
;
int vis
,head
,match
;
bool crosspath(int u)
{
for(int i = 0;i<G[u].size() ; i++)
{
int v=G[u][i];
if(!vis[v])
{
vis[v]=1;
if(match[v]==-1||crosspath(match[v]))
{
match[v]=u;
return true;
}
}
}
return false;
}
int n,tot;
int hungary()
{
int ans=0;
for(int i = 0;i <= n-1; i++)
{
cle(vis);
if(crosspath(i))ans++;
}
return ans/2;
}
int main()
{
freopen(FROP".in","r",stdin);
freopen(FROP".out","w",stdout);
while(scanf("%d",&n)==1)
{
int tot=0;
cle(match);
for(int i = 1; i <= n; i++)
{
int x;
scanf("%d",&x);
getchar();getchar();
int w;
scanf("%d",&w);
getchar();
for(int j= 1; j <= w; j++)
{
int y;
scanf("%d",&y);
G[x].push_back(y);
G[y].push_back(x);
}
}
mincle(match);
printf("%d\n",hungary());
for(int i = 0; i< n; i++)
G[i].clear();
}
return 0;
}