c/c++语言开发共享洛谷 P2756 飞行员配对方案问题

#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("%dn",hungry());          for(int i = 1;i <= m;++i)         if(to[i]) printf("%d %dn",i,to[i]);     return 0; } 

//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("%dn",dinic());     for(int i = l;i < r;i += 2){         if(edge[i].w == 0){             printf("%d %dn",edge[i^1].to,edge[i].to);         }     }     return 0; }