HDU 1150 Machine Schedule 二分图最小覆盖点

C - Machine Schedule

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

As we all know, machine scheduling is a very classical problem in computer science and has been studied for a very long history. Scheduling problems differ widely in the nature of the constraints that must be satisfied and the type of schedule desired. Here we consider a 2-machine scheduling problem. 

There are two machines A and B. Machine A has n kinds of working modes, which is called mode_0, mode_1, ..., mode_n-1, likewise machine B has m kinds of working modes, mode_0, mode_1, ... , mode_m-1. At the beginning they are both work at mode_0. 

For k jobs given, each of them can be processed in either one of the two machines in particular mode. For example, job 0 can either be processed in machine A at mode_3 or in machine B at mode_4, job 1 can either be processed in machine A at mode_2 or in machine B at mode_4, and so on. Thus, for job i, the constraint can be represent as a triple (i, x, y), which means it can be processed either in machine A at mode_x, or in machine B at mode_y. 

Obviously, to accomplish all the jobs, we need to change the machine's working mode from time to time, but unfortunately, the machine's working mode can only be changed by restarting it manually. By changing the sequence of the jobs and assigning each job to a suitable machine, please write a program to minimize the times of restarting machines. 

Input

The input file for this program consists of several configurations. The first line of one configuration contains three positive integers: n, m (n, m < 100) and k (k < 1000). The following k lines give the constrains of the k jobs, each line is a triple: i, x, y. 

The input will be terminated by a line containing a single zero. 

Output

The output should be one integer per line, which means the minimal times of restarting machine.

Sample Input

5 5 100 1 11 1 22 1 33 1 44 2 15 2 26 2 37 2 48 3 39 4 30

Sample Output

3

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 1005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;int bmap[maxn][maxn];bool bmask[maxn];int nx,ny,m;int pre[maxn];int findpath(int u){    FOR(i,1,ny)        if(bmap[u][i]&&!bmask[i]){                bmask[i]=1;                if(pre[i]==-1||findpath(pre[i])){                    pre[i]=u;                    return 1;                }        }    return 0;}int MaxMatch(){    int res(0);    FOR(i,1,nx){            MT(bmask,0);            res+=findpath(i);        }    return res;}void init(){    MT(bmap,0);    MT(bmask,0);    MT(pre,-1);}int main(){    while(rd(nx)&&nx){        rd2(ny,m);        int x,y,k;init();        FOR(i,1,m){            rd3(k,x,y);            if(x+y)                bmap[x][y]=1;        }        printf("%d\n",MaxMatch());    }    return 0;}/*5 5 100 1 11 1 22 1 33 1 44 2 15 2 26 2 37 2 48 3 39 4 30*/