HDU 1054 Strategic Game（二分图 匈牙利算法）

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

40:(1) 11:(2) 2 32:(0)3:(0)53:(3) 1 4 21:(1) 02:(0)0:(0)4:(0)

 

Sample Output

12

 

题意：求最少的点覆盖所有的边，即最少的点与所有的边有关

思路：最小点覆盖=最大匹配数

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b)  for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 1505vector<int>g[N];int n;int link[N],vis[N];int ans;bool dfs(int x){    int i,j;    fre(i,0,g[x].size())    {        if(vis[g[x][i]]) continue;        vis[g[x][i]]=1;        if(link[g[x][i]]==-1||dfs(link[g[x][i]]))        {            link[g[x][i]]=x;            return true;        }    }   return false;}void solve(){    int i,j;    mem(link,-1);    ans=0;    fre(i,0,n)    {       mem(vis,0);       if(dfs(i)) ans++;    }  pf("%d\n",ans/2);}int main(){    int i,j;    while(~sf(n))    {        fre(i,0,n)         g[i].clear();        int x,y,t;        fre(i,0,n)        {            scanf("%d:(%d)",&x,&t);            while(t--)            {              sf(y);              g[x].push_back(y);              g[y].push_back(x);            }        }       solve();    }   return 0;}