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TOJ 3345 Chinese Chess 二分匹配 匈牙利算法 扩展

TOJ 3345 Chinese Chess http://acm.tju.edu.cn/toj/showp3345.html
双链表实习双向邻接表,正向图反向图,匈牙利匹配后,枚举删除匹配边,两边做重新查找增广路操作,如果都没法查到则证明是重要边,还原原匹配时候要注意细节。

优化1:求最大匹配时候先能匹配就匹配上。
优化2:标记已经确定不是必要点的行和列 flag[]和flag2[]
注意1:删除匹配边后 应在匹配边的两面点 分别查询
注意2:没查到新匹配要还原匹配边

#include
<stdio.h>
#include
<string.h>
const int size2 = 100110;
const int size1 = 10110;
int n , m , k ;
struct node{
    node 
*p;
    node 
*q;
    
int v;
    
int v2;
};
int val;

node edg[size2];

bool visit[size1];
int pp[size1];
bool flag[size1];
bool flag2[size1];

node 
*g[size1];
node 
*g1[size1];

int ppre[size1];
int ppr[size1];

int nxy[size1];
inline 
bool check(int x , int y , int fc){
    node 
*now = g[x];
    
while(now != NULL){
        
if(now->== y){
            
if(fc == 0)
                
return true;
            
if(fc == 1){
                now
->= -1;
                
return true;
            }
            
if(fc == 2){
                now
->= val;
                
return true;
            }
        }
        
else{
            now 
= now->p;
        }
    }
    
return false;
}

inline 
bool check2(int x , int y , int fc){
    node 
*now = g1[x];
    
while(now != NULL){
        
if(now->v2 == y){
            
if(fc == 0)
                
return true;
            
if(fc == 1){
                now
->v2 = -1;
                
return true;
            }
            
if(fc == 2){
                now
->v2 = val;
                
return true;
            }
        }
        
else{
            now 
= now->q;
        }
    }
    
return false;
}
inline 
bool find(int a){
    node 
*now = g[a];
    
while(now != NULL){
             
int i = now->v;
             
if(i != -1){
                 
if(!visit[i] ){
                    visit[i] 
= true;
                    
if(pp[i] == -1 ||find(pp[i])){
                        pp[i] 
= a;
                        nxy[a] 
= i;
                        
return true;
                    }
                }
             }
            now 
= now->p;
    }
    
return false;
}

inline 
bool find2(int a){
    node 
*now = g1[a];
    
while(now != NULL){
             
int i = now->v2;
             
if(i != -1){
                 
if(!visit[i] ){
                    visit[i] 
= true;
                    
if(nxy[i] == -1 ||find2(nxy[i])){
                        nxy[i] 
= a;
                        pp[a] 
= i;
                        
return true;
                    }
                }
             }
            now 
= now->q;
    }
    
return false;
}

void output(){
    printf(
"\n\n");
    
forint i = 1 ; i <= m ; i++)
        printf(
"%d -> %d\n",pp[i],i);
        printf(
"\n");
    
forint i = 1 ; i <= n ; i++)
        printf(
"%d -> %d\n",i,nxy[i]);
    printf(
"\n\n");
}

void output1(int x){
    node 
*now = g1[x];
    
while(now != NULL){
             
int i = now->v2;

             printf(
"  %d",i);
            now 
= now->q;
    }
    printf(
"\n");
}

int main(){
        
int a1,i,a2;
        
int _case = 0;
        
int num;
        
while(scanf("%d%d%d",&n,&m,&k)!=EOF){
            num 
= 0;
            memset(g,
0,sizeof(g));
            memset(g1,
0,sizeof(g1));
            memset(pp,
-1,sizeof(pp));
            memset(nxy,
-1,sizeof(nxy));
            
for(i = 1; i <= k ; i++){
                scanf(
"%d%d",&a1,&a2);
                
if(check(a1,a2,0)) continue;
                edg[i].v 
= a2;
                edg[i].v2 
= a1;
                edg[i].p 
= NULL;
                edg[i].q 
= NULL;

                
if(g[a1] == NULL){
                    g[a1] 
= &edg[i];
                }
else{
                    edg[ppre[a1]].p 
= &edg[i];
                }

                
if(g1[a2] == NULL){
                    g1[a2] 
= &edg[i];
                }
else{
                    edg[ppr[a2]].q 
= &edg[i];
                }

                ppre[a1] 
= i ;
                ppr[a2] 
= i ;

                
if(pp[a2] == -1 && nxy[a1] == -1){
                    pp[a2] 
= a1;
                    nxy[a1] 
= a2;
                    num
++;
                }
            }
            
for(i = 1 ; i <= m ; i++){
                
if(pp[i] == -1){
                    memset(visit,
0,sizeof(visit));
                    
if(find2(i)){
                        num
++;
                    }
                }
            }
            memset(flag,
0,sizeof(flag));
            memset(flag2,
0,sizeof(flag2));
            
int ans = 0;
            
for(i = 1 ; i <= m ; i++){
                
if(pp[i] != -1){
                    
int key = 0;
                    
int now = pp[i];
                    pp[i] 
= -1 ;
                    nxy[now] 
= -1;
                    
if(flag[pp[i]] == 0){
                        check(now,i,
1);
                        memset(visit,
0,sizeof(visit));
                        
if(find(now)){
                            key 
= 1;
                        }
                        val 
= i ;
                        check(now,
-1,2);
                    }

                    
if(flag2[i] == 0 && key == 0 ){
                        check2(i,now,
1);
                        memset(visit,
0,sizeof(visit));
                        
if(find2(i)){
                            key 
= 1;
                        }
                        val 
= now ;
                        check2(i,
-1,2);
                    }
                     
if(!key){
                            ans
++;
                                nxy[now] 
= i;
                            pp[i] 
= now;
                     }
                     
else{
                        flag[now] 
= 1 ;
                        flag2[i] 
= 1 ;
                     }
                }
//右侧有边
            }
            printf(
"Board %d have %d important blanks for %d chessmen.\n",++_case,ans,num);
        }
    
return 0;
}

posted on 2009-08-07 00:17 hadn't 阅读(303) 评论(0)  编辑 收藏 引用


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