Hdu1054--strategic Game (minimum vertex overlay + adjacency table)

Problem Description
Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and T Hen 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 in the nodes so, they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob have 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 is 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 a soldier (at the Node 1).

The output should is printed on the standard output. For each given input data set, print one, integer number in a, gives the result (the minimum number of sold Iers). An example are 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

With the fewest points to defend the surrounding points, conversion to a binary map is the smallest vertex coverage problem, of course, the premise is to understand the conversion ...
In addition, the data is a bit large, can not use the adjacency matrix, but poj on the adjacency table can not be used.

#include <iostream> #include <cstring> #include <string> #include <vector> #include <queue > #include <cstdio> #include <set> #include <cmath>//#include <map> #include <algorithm&
Gt
#define INF 0x3f3f3f3f #define MAXN 1505 #define MOD 10001 using namespace std;
BOOL VIS[MAXN];  int pre[maxn],n;
Matching path;
Vector<int> MAP[MAXN];
        int find (int cur) {for (int i=0; i<map[cur].size (); ++i) {int v=map[cur][i];
            if (!vis[v]) {Vis[v] = true;
                if (pre[v] = = 0 | | find (PRE[V)) {pre[v] = cur;
            return 1;
}}} return 0;
        } int main () {while (~SCANF ("%d", &n)) {memset (map,0,sizeof (map));
        int u,v,m;
            for (int i=0;i<n;++i) {scanf ("%d: (%d)", &u,&m);
                while (m--) {scanf ("%d", &v); Map[v].push_back(u);
            Map[u].push_back (v);
        }} int ans=0;
        memset (pre,0,sizeof (pre));
            for (int i=0;i<n;++i) {memset (vis,0,sizeof (VIS));
        if (find (i)) ans++;
    } printf ("%d\n", ANS/2);
} return 0;
 }