URAL 1774 Barber of the Army of Mages (最大流)

将源点和每个时间点连一条容量为k的边，表示同一时间最多剃k次胡子，

将一个人和对应时间连一条容量为1的边，表示一个人在某个时间只能剃1次，

再将每个人和汇点连一条容量为2的边表示一个人要剃两次。

先是想贪心，结果不对，后来想到网络流，不会写好伤，下来写了好几发才过，写得依然丑。

对最大流的理解加深了，图论有待加强

（似乎贪心对次数不能很好处理

#include<bits/stdc++.h>
using namespace std;

const int INF = 0x3fffffff;
const int maxn = 2142;
#define PB push_back
struct Edge
{
    int from,to,cap,flow;
};

vector<Edge> edges;
vector<int> G[maxn];
int n,m,S = 2132,T = 2133;
int Hum = 2002;
void AddEdge(int from,int to,int cap)
{
    edges.PB(Edge{from,to,cap,0});
    edges.PB(Edge{to,from,0,0});
    int sz = edges.size();
    G[from].PB(sz-2);
    G[to].PB(sz-1);
}

bool vis[maxn];
int d[maxn],cur[maxn];

bool bfs()
{
    memset(vis,0,sizeof(vis));
    queue<int> q;
    q.push(S);
    d[S] = 0;
    vis[S] = true;
    while(q.size()){
        int u = q.front(); q.pop();
        for(int i = 0; i < G[u].size(); i++){
            Edge &e = edges[G[u][i]];
            if(!vis[e.to] && e.cap > e.flow){
                vis[e.to] = true;
                d[e.to] = d[u] + 1;
                q.push(e.to);
            }
        }
    }
    return vis[T];
}

int dfs(int u,int a)
{
    if(u == T || a == 0) return a;
    int flow = 0, f;
    for(int &i = cur[u]; i < G[u].size(); i++){
        Edge &e = edges[G[u][i]];
        if(d[u] + 1 == d[e.to] && (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 ans[101][2];

int main()
{
    //freopen("in.txt","r",stdin);
    int N,k; scanf("%d%d",&N,&k);
    int Se = INF,Ed = -INF;
    for(int i = 0; i < N; i++){
        int s,w; scanf("%d%d",&s,&w);
        w += s;
        Se = min(Se,s);
        Ed = max(Ed,w);
        for(int j = s; j < w; j++){
            AddEdge(j,Hum+i,1);
        }
    }
    for(int i = Se; i < Ed; i++){
        AddEdge(S,i,k);
    }
    for(int i = Hum,maxi = Hum+N; i < maxi; i++){
        AddEdge(i,T,2);
    }
    int flow = MaxFlow();
    if(flow != 2*N){
        puts("No");
    }else {
        puts("Yes");
        for(int i = Hum,maxi = Hum+N; i < maxi; i++){
            int cnt = 0;
            for(int j = 0; j < G[i].size()&& cnt < 2; j++){
                Edge & e = edges[G[i][j]];
                if(e.cap == 0 && e.flow == -1){
                    ans[i-Hum][cnt++] = e.to;
                }
            }
        }
        for(int i = 0; i < N; i++){
            printf("%d %d
",ans[i][0],ans[i][1]);
        }
    }
    return 0;
}