洛谷 P2756 飞行员配对方案问题

飞行员配对方案问题(求最大匹配数并且输出配对方案)

　　两种做法:

　　1)二分图匹配匈牙利算法,可以直接求出最大匹配数,并且数组中记录了最佳配对方案

　　2)最大流,超级源点S到A集合中每一个元素建边(容量为1),B集合中每一个点到汇点建边(容量为1),A集合中与B集合中可匹配的点之间建边(容量为1),跑最大流即可得到最大匹配数.

　　由于Dinic算法中被使用过的路流量会减少,而它的反向边流量会增加.

　　所以在最大匹配的时候,AB集合中两个点若配对,有 A -> B的边容量为0 (或B -> A的边容量为1).

 

二分图匈牙利版本：

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#include <bits/stdc++.h> using namespace std; #define N 105 #define M 20005 struct Edge{     int to,next; }edge[M]; int n,m; int vis[N],match[N],to[N],head[N];//初始化match -> -1 ,to -> 0 int cnt; void init(){     cnt = 0;     memset(head,-1,sizeof(head));     memset(match,-1,sizeof(match));     memset(to,0,sizeof(to)); }   bool dfs(int u){     for(int i = head[u];i != -1;i = edge[i].next){         int v = edge[i].to;         if(!vis[v]){             vis[v] = 1;             if(match[v] == -1 || dfs(match[v])){                 match[v] = u;                 to[u] = v;                 return true;             }         }     }     return false; } //cnt 为最多匹配到的点数(单边点数) int hungry(){     cnt = 0;     for(int i = 1;i <= n;++i){         memset(vis,0,sizeof(vis));         if(!to[i]) cnt += dfs(i);     }     return cnt; }   int main() {     init();     scanf("%d%d",&m,&n);     int u,v;     while(scanf("%d%d",&u,&v) && u != -1){         edge[cnt].to = v,edge[cnt].next = head[u],head[u] = cnt++;     }     printf("%d\n",hungry());           for(int i = 1;i <= m;++i)         if(to[i]) printf("%d %d\n",i,to[i]);     return 0; }

 

最大流dinic版本：

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//最大流Dinic算法 //m为边数,n为点数 //复杂度O(m*n*n) #include <bits/stdc++.h> using namespace std; #define N 105 #define M 20005 int INF = 0x3f3f3f3f; int dep[N],head[N]; int to[N];//当前弧优化 int n,m; struct Edge{     int to,next,w; }edge[M<<1]; int cnt = 0; //edge[i] 的反向边为 dege[i^1]   int s,t;//s->源,t->汇   void ad(int x,int y,int w){     edge[cnt].to = y,edge[cnt].next = head[x],edge[cnt].w = w,head[x] = cnt++;     edge[cnt].to = x,edge[cnt].next = head[y],edge[cnt].w = 0,head[y] = cnt++; } void init(){     memset(to,0,sizeof(to));     memset(head,-1,sizeof(head));     cnt = 0; }   bool D_bfs(){     memset(dep,0,sizeof(dep));     memset(to,0,sizeof(to));     dep[s] = 1;     queue<int> Q;     while(!Q.empty()) Q.pop();     Q.push(s);     while(!Q.empty()){         int u = Q.front();         Q.pop();         for(int i = head[u];i != -1;i = edge[i].next){             int v = edge[i].to;             if(edge[i].w > 0 && !dep[v]){                 dep[v] = dep[u] + 1;                 Q.push(v);             }         }     }     if(dep[t]) return 1;     return 0; }   int D_dfs(int u,int now){     if(u == t) return now;     int beg = to[u] ? to[u] : head[u];     for(int i = beg;i != -1;i = edge[i].next){         int v = edge[i].to;         if(dep[v] == dep[u] + 1 && edge[i].w > 0){             int di = D_dfs(v,min(now,edge[i].w));             if(di == 0) continue;             edge[i].w -= di;             edge[i^1].w += di;             if(edge[i].w) to[u] = i;             else to[u] = edge[i].next;             return di;         }     }     return 0; }   int Dinic(){     int sum = 0,flow;     while(D_bfs()){         while((flow = D_dfs(s,INF)))             sum += flow;     }     return sum; } int l , r; void getMap(){     scanf("%d%d",&m,&n);     s = 0,t = n+1;     for(int i = 1;i <= m;++i) ad(0,i,1);     for(int i = m+1;i <= n;++i) ad(i,n+1,1);     int u,v;     l = cnt;     while(scanf("%d%d",&u,&v) && u != -1) ad(u,v,1);     r = cnt; }   int main() {     init();     getMap();     printf("%d\n",Dinic());     for(int i = l;i < r;i += 2){         if(edge[i].w == 0){             printf("%d %d\n",edge[i^1].to,edge[i].to);         }     }     return 0; }