洛谷 P2756 飞行员配对方案问题

全名:线性规划与网络流24题

按照题目难度顺序:

1.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：

#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：

//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;
}