HDU1054（二分图）

Strategic Game

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6941 Accepted Submission(s): 3263

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

因为给出的是一棵树，所以他一定可转化成一个二分图，那么问题就转化成求二分图的最小点覆盖（最小点覆盖：假如选了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点来覆盖所有的边。），利用König定理，它的意思是，一个二分图中的最大匹配数等于这个图中的最小点覆盖数。知道这个就变成模版题了，但是因为点比较多所以要用Hopcroft-Carp算法优化一波，试过用普通匈牙利算法TLE了。

/* ********************************************* 二分图匹配（Hopcroft-Carp的算法）。 初始化：g[][]邻接矩阵 调用：res=MaxMatch(); Nx,Ny要初始化！！！ 时间复杂大为 O(V^0.5 E) 适用于数据较大的二分匹配 需要ｑｕｅｕｅ头文件 ********************************************** */
#include "cstring"
#include "cstdio"
#include "string.h"
#include "iostream"
#include "queue"
using namespace std;
const int MAXN=1510;
const int INF=1<<28;
int g[MAXN][MAXN],Mx[MAXN],My[MAXN],Nx,Ny;
int dx[MAXN],dy[MAXN],dis;
bool vst[MAXN];
bool searchP()
{
    queue<int>Q;
    dis=INF;
    memset(dx,-1,sizeof(dx));
    memset(dy,-1,sizeof(dy));
    for(int i=0;i<Nx;i++)
        if(Mx[i]==-1)
        {
            Q.push(i);
            dx[i]=0;
        }
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        if(dx[u]>dis)  break;
        for(int v=0;v<Ny;v++)
            if(g[u][v]&&dy[v]==-1)
            {
                dy[v]=dx[u]+1;
                if(My[v]==-1)  dis=dy[v];
                else
                {
                    dx[My[v]]=dy[v]+1;
                    Q.push(My[v]);
                }
            }
    }
    return dis!=INF;
}
bool DFS(int u)
{
    for(int v=0;v<Ny;v++)
        if(!vst[v]&&g[u][v]&&dy[v]==dx[u]+1)
        {
            vst[v]=1;
            if(My[v]!=-1&&dy[v]==dis) continue;
            if(My[v]==-1||DFS(My[v]))
            {
                My[v]=u;
                Mx[u]=v;
                return 1;
            }
        }
    return 0;
}
int MaxMatch()
{
    int res=0;
    memset(Mx,-1,sizeof(Mx));
    memset(My,-1,sizeof(My));
    while(searchP())
    {
        memset(vst,0,sizeof(vst));
        for(int i=0;i<Nx;i++)
            if(Mx[i]==-1&&DFS(i))  res++;
    }
    return res;
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(g,0,sizeof(g));
        Nx=Ny=n;
        for(int i=1;i<=n;i++)
        {
            int now,cnt;
            scanf("%d:(%d)",&now,&cnt);
            while(cnt--)
            {
                int temp;
                scanf("%d",&temp);
                g[now][temp]=1;
                g[temp][now]=1;
            }
        }
        printf("%d\n",MaxMatch()/2);
    }
}