【loj】#6006. 「网络流 24 题」试题库(二分图匹配)

记录一个菜逼的成长。。

题目链接

这题跟前面那到圆桌聚餐差不多。

#include <bits/stdc++.h>using namespace std;#define rep(i,l,r) for( int i = l; i <= r; i++ )#define rep0(i,l,r) for( int i = l; i < r; i++ )#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define clr clear()#define pb push_back#define mp make_pair#define fi first#define se secondtypedef long long LL;typedef pair<int,int> PII;const int INF = 0x3f3f3f3f;const int MAX_V = 2000 + 10;struct edge{  int to,cap,rev,flow;  edge(){}  edge(int _to,int _cap,int _rev,int _flow):to(_to),cap(_cap),rev(_rev),flow(_flow){}};vector<edge>G[MAX_V];int level[MAX_V];int iter[MAX_V];void add(int from,int to,int cap,int flow = 0){  G[from].push_back(edge(to,cap,G[to].size(),0));  G[to].push_back(edge(from,0,G[from].size()-1,0));}void bfs(int s){  memset(level,-1,sizeof(level));  queue<int>que;  level[s] = 0;  que.push(s);  while(!que.empty()){    int f = que.front();    que.pop();    for( int i = 0; i < G[f].size(); i++ ){      edge &e = G[f][i];      if(e.cap > 0 && level[e.to] == -1){        level[e.to] = level[f] + 1;        que.push(e.to);      }    }  }}int dfs(int v,int t,int f){  if(v == t)return f;  for( int &i = iter[v]; i < G[v].size(); i++ ){    edge &e = G[v][i];    if(e.cap > 0 && level[v] < level[e.to]){      int d = dfs(e.to,t,min(e.cap,f));      if(d > 0){        e.cap -= d;e.flow += d;        G[e.to][e.rev].cap += d;        return d;      }    }  }  return 0;}int max_flow(int s,int t){  int flow = 0;  for(;;){    bfs(s);    if(level[t] == -1)return flow;    memset(iter,0,sizeof(iter));    int f;    while((f = dfs(s,t,INF)) > 0)      flow += f;  }}int main(){  int k,n;  scanf("%d%d",&k,&n);  int s = 0,t = n + k + 1,sum = 0;  rep(i,1,k){    int x;scanf("%d",&x);sum += x;    add(i+n,t,x);  }  rep(i,1,n){    int p;scanf("%d",&p);    rep(j,1,p){      int x;scanf("%d",&x);      add(i,x+n,1);    }  }  rep(i,1,n){    add(s,i,1);  }  int ret = max_flow(s,t);  if(ret < sum){    puts("No Solution!");    return 0;  }  vector<int>a[MAX_V];  rep(i,1,n){    for( int j = 0; j < G[i].size(); j++ ){      if(G[i][j].to > n && G[i][j].cap == 0){        a[G[i][j].to-n].pb(i);      }    }  }  rep(i,1,k){    printf("%d: ",i);    for( int j = 0; j < a[i].size(); j++ ){      printf("%d ",a[i][j]);    }    puts("");  }  return 0;}