HDU1054 (two-part picture)

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 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

Source
Southeastern Europe 2000

Because the given is a tree, so he must be converted into a binary map, then the problem is converted to a binary map of the minimum point coverage (minimum point coverage: If you choose a point is equivalent to covering all the edges with it as the end point, you need to select the fewest points to cover all the edges. ), using the K?nig theorem, which means that the maximum number of matches in a binary graph is equal to the minimum number of points covered in the graph. Know this becomes a template problem, but because the point is more so to use the HOPCROFT-CARP algorithm to optimize a wave, tried to use the ordinary Hungarian algorithm tle.

/* ********************************************* binary graph matching (hopcroft-carp algorithm).  Initialize: g[][] adjacency matrix call: Res=maxmatch (); Nx,ny to initialize!!! Time is complicated. O (v^0.5 E) suitable for large binary matches requires the queue header file ********************************************** */#include "CString"#include "Cstdio"#include "string.h"#include "iostream"#include "queue"using namespace STD;Const intmaxn=1510;Const intinf=1<< -;intG[maxn][maxn],mx[maxn],my[maxn],nx,ny;intDx[maxn],dy[maxn],dis;BOOLVST[MAXN];BOOLSEARCHP () { Queue<int>Q; Dis=inf;memset(dx,-1,sizeof(DX));memset(dy,-1,sizeof(DY)); for(intI=0; i<nx;i++)if(mx[i]==-1) {Q.push (i); dx[i]=0; } while(! Q.empty ()) {intU=q.front (); Q.pop ();if(Dx[u]>dis) Break; for(intv=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]); }            }    }returnDis!=inf;}BOOLDFS (intu) { for(intv=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;}intMaxmatch () {intres=0;memset(mx,-1,sizeof(Mx));memset(my,-1,sizeof(My)); while(SEARCHP ()) {memset(VST,0,sizeof(VST)); for(intI=0; i<nx;i++)if(mx[i]==-1&&dfs (i)) res++; }returnRes;}intMain () {intN while(~scanf("%d", &n)) {memset(g,0,sizeof(g)); Nx=ny=n; for(intI=1; i<=n;i++) {intnow,cnt;scanf("%d: (%d)", &now,&cnt); while(cnt--) {intTempscanf("%d", &temp); g[now][temp]=1; g[temp][now]=1; }        }printf("%d\n", Maxmatch ()/2); }}

HDU1054 (two-part picture)