Bipartite graph matching example code and collation

2. Sample code matching and sorting

1. Hungarian algorithm

HDU 1150

#include<stdio.h> 
#include<string.h> 
#include<algorithm> 
using namespace std; 
int m,n,k; 
int vis[105]; 
int mpt[105][105]; 
int use[105]; 
int hungary(int x) 
{ 
  for(int i=1;i<m;i++) 
  { 
    if(vis[i]==0&&mpt[x][i]==1) 
    { 
      vis[i]=1; 
      if(use[i]==-1||hungary(use[i])) 
      { 
        use[i]=x; 
        return 1; 
      } 
    } 
  } 
  return 0; 
} 
int main() 
{ 
  while(scanf("%d",&n)!=EOF,n) 
  { 
    scanf("%d%d",&m,&k); 
    int a,b,c; 
    memset(mpt,0,sizeof(mpt)); 
    for(int i=1;i<=k;i++) 
    { 
      scanf("%d%d%d",&c,&a,&b); 
      mpt[a][b]=1; 
    } 
    int ans=0; 
    memset(use,-1,sizeof(use)); 
    for(int i=1;i<n;i++) 
    { 
      if(hungary(i)) 
      { 
        ans++; 
      } 
      memset(vis,0,sizeof(vis)); 
    } 
    printf("%d\n",ans); 
  } 
  return 0; 
} 

2. KM algorithm

HDU 2255

Read a lot of materials are not very understand, the first post others' template

#include<iostream> 
#include<cstdio> 
#include<cstring> 
#include<climits> 
#include<algorithm> 
using namespace std; 
#define N 310 
int map[N][N]; 
bool visitx[N], visity[N]; 
int lx[N], ly[N]; 
int match[N]; 
int n; 
 
bool Hungary(int u) // Hungarian algorithm  
{ 
  visitx[u] = true; 
  for(int i = 0; i < n; ++i) 
  { 
    if(!visity[i] && lx[u] + ly[i] == map[u][i]) 
    { 
      visity[i] = true; 
      if(match[i] == -1 || Hungary(match[i])) 
      { 
        match[i] = u; 
        return true; 
      } 
    } 
  } 
  return false; 
} 
 
void KM_perfect_match() 
{ 
  int temp; 
  memset(lx, 0, sizeof(lx)); // Initializes the header  
  memset(ly, 0, sizeof(ly)); //ly[i] for 0 
  for(int i = 0; i < n; ++i) //lx[i] Is the edge with the largest weight  
    for(int j = 0; j < n; ++j) 
      lx[i] = max(lx[i], map[i][j]); 
  for(int i = 0; i < n; ++i) // right n A point matching  
  { 
    while(1) 
    { 
      memset(visitx, false, sizeof(visitx)); 
      memset(visity, false, sizeof(visity)); 
      if(Hungary(i)) // The match is successful  
        break; 
      else // The match failed. Find the minimum  
      { 
        temp = INT_MAX; 
        for(int j = 0; j < n; ++j) //x In a staggered tree  
          if(visitx[j]) 
            for(int k = 0; k < n; ++k) //y Outside the staggered tree  
              if(!visity[k] && temp > lx[j] + ly[k] - map[j][k]) 
                temp = lx[j] + ly[k] - map[j][k]; 
        for(int j = 0; j < n; ++j) // Update the top mark  
        { 
          if(visitx[j]) 
            lx[j] -= temp; 
          if(visity[j]) 
            ly[j] += temp; 
        } 
      } 
    } 
  } 
} 
 
int main() 
{ 
  int ans; 
  while(scanf("%d", &n) != EOF) 
  { 
    ans = 0; 
    memset(match, -1, sizeof(match)); 
    for(int i = 0; i < n; ++i) 
      for(int j = 0; j < n; ++j) 
        scanf("%d", &map[i][j]); 
    KM_perfect_match(); 
    for(int i = 0; i < n; ++i) // A weight added  
      ans += map[match[i]][i]; 
    printf("%d\n", ans); 
  } 
  return 0; 
} 

3. Multiple matching

HDU 3605 Escape

#include<stdio.h> 
#include<string.h> 
#include<algorithm> 
using namespace std; 
int n,m; 
int num[15]; 
int mpt[100005][15]; 
int vis[15]; 
int use[15]; 
int dp[15][100005]; 
int hungary(int x) 
{ 
  for(int i=1;i<=m;i++) 
  { 
    if(vis[i]==0&&mpt[x][i]==1) 
    { 
      vis[i]=1; 
      if(use[i]<num[i])// Meet the conditions  
      { 
        dp[i][use[i]++]=x; 
        return 1; 
      } 
      // If you're not satisfied, look for an augmented path  
      for(int j=0;j<use[i];j++)// See if you can go back 1 A go out  
      { 
        if(hungary(dp[i][j])) 
        { 
          dp[i][j]=x; 
          return 1; 
        } 
      } 
    } 
  } 
  return 0; 
} 
int main() 
{ 
  while(scanf("%d%d",&n,&m)!=EOF) 
  { 
    for(int i=1;i<=n;i++) 
    { 
      for(int j=1;j<=m;j++) 
      { 
        scanf("%d",&mpt[i][j]); 
      } 
    } 
    for(int i=1;i<=m;i++) 
      scanf("%d",&num[i]); 
    int ans=0; 
    memset(use,0,sizeof(use)); 
    for(int i=1;i<=n;i++) 
    { 
      memset(vis,0,sizeof(vis)); 
      if(!hungary(i)) 
      { 
        ans=1; 
        break; 
      } 
    } 
    if(ans==0) 
    { 
      printf("YES\n"); 
    } 
    else printf("NO\n"); 
  } 
 
  return 0; 
} 

The above is 2 sub map matching code, if you have questions please leave a message, or to the site community exchange discussion, thank you for reading, hope to help you, thank you for the support of the site!