1736: 飞行员配对方案问题 二分最大匹配/最大流

假设有10个驾驶员，如图中的V1，V2，…，V10就代表达10个驾驶员,其中V1，V2，V3，V4，V5是正驾驶员,V6，V7，V8，V9，V10是副驾驶员。

如果一个正驾驶员和一个副驾驶员可以同机飞行，就在代表他们两个之间连一条线,两个人不能同机飞行，就不连。例如V1和V7可以同机飞行，而V1和V8就不行。

请搭配飞行员，使出航的飞机最多。注意:因为驾驶工作分工严格,两个正驾驶员或两个副驾驶员都不能同机飞行.

二分最大匹配版

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 100 + 5;
vector<int> G[maxn];
int link[maxn];
bool used[maxn];
int n,m;
void Init()
{
    for(int i = 0;i < maxn;++i)
        G[i].clear();
}
void AddEdge(int from,int to)
{
   G[from].push_back(to);
}
bool dfs(int u)
{
    for(int i = 0;i < (int)G[u].size();++i){
        int v = G[u][i];
        if(!used[v]){
            used[v] = true;
            if(link[v]==-1||dfs(link[v])){
                link[v] = u;      //printf("%d ---> %d
",u,v);
                return true;
            }
        }
    }
    return false;
}
int Find()
{
    int res = 0;
    memset(link,-1,sizeof(link));
    for(int i = 1;i <= n;++i){
        memset(used,false,sizeof(used));
        if(dfs(i))res++;
    }
    return res;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int x,y;
        Init();
        while(scanf("%d%d",&x,&y),(x!=-1&&y!=-1)){
            AddEdge(x,y);
        }
        int res = Find();
        if(!res)
            puts("No Solution!");
        else{
            printf("%d
",res);
            for(int i = n+1;i <= m;++i){
              if(link[i]!=-1)
                printf("%d %d
",link[i],i);
            }
        }
    }
    return 0;
}

网络流最大流

#include <iostream>
#include <vector>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 100+5;
const int INF = 200;
struct Edge{
   int from,to,cap,flow;
};
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn],cur[maxn];
bool vst[maxn];
int n,m,s,t;
void Init()
{
    for(int i = 0;i < maxn;++i)
        G[i].clear();
    edges.clear();
}
void AddEdge(int from,int to,int cap)
{
    edges.push_back((Edge){from,to,cap,0});
    edges.push_back((Edge){to,from,0,0});
    int sz = edges.size();
    G[from].push_back(sz-2);
    G[to].push_back(sz-1);
}
//构造层次图
bool BFS()
{
    memset(vst,false,sizeof(vst));
    queue<int> Q;
    Q.push(s);
    d[s] = 0;
    vst[s] = true;
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        for(int i = 0;i < (int)G[u].size();++i){
            Edge& e = edges[G[u][i]];
            if(!vst[e.to]&&e.cap > e.flow){
                d[e.to] = d[u]+1;
                vst[e.to] = true;
                Q.push(e.to);
            }
        }
    }
    return vst[t];
}
//增广
int DFS(int u,int a)
{
    if(u==t||a==0)
        return a;
    int f,flow = 0;
    for(int& i = cur[u];i < (int)G[u].size();++i){
        Edge& e = edges[G[u][i]];
        if(d[e.to]==d[u]+1&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
            e.flow += f;
            edges[G[u][i]^1].flow -= f;
            flow += f;
            a -= f;
            if(a==0)break;
        }
    }
    return flow;
}
int Maxflow()
{
    int flow = 0;
    while(BFS()){
        memset(cur,0,sizeof(cur));
        flow += DFS(s,INF);
    }
    return flow;
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        Init();
        int x,y;
        while(scanf("%d%d",&x,&y),(x!=-1&&y!=-1)){
            AddEdge(x,y,1);
//            AddEdge(y,x,1);
        }
        s = 0,t = m+1;
        for(int i = 1;i <= n;++i)AddEdge(s,i,1);
        for(int i = n+1;i <= m;++i)AddEdge(i,t,1);
        int res = Maxflow();
        if(res<=0)
            puts("No Solution!");
        else{
            printf("%d
",res);
            int sz = edges.size();
            for(int i = 0;i < sz;i += 2){
                Edge& e = edges[i];
                if(e.flow == 1&&e.from!=s&&e.to!=t)
                    printf("%d %d
",e.from,e.to);
            }
        }
    }
    return 0;
}