【loj】#6004. 「网络流 24 题」圆桌聚餐(二分图匹配)

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记录一个菜逼的成长。。

题目链接

二分图匹配，由于一个代表只能一桌，同一个代表团不能坐同一桌
所以代表团跟桌的流量设为1

#include <bits/stdc++.h>using namespace std;#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 = 1000 + 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 vis[MAX_V];int main(){  int n,m;  scanf("%d%d",&n,&m);  int s = 0,t = n + m + 1;  int sum = 0;  for( int i = 1; i <= n; i++ ){    int x;scanf("%d",&x);    sum += x;    add(s,i,x);  }  for( int i = 1; i <= n; i++ ){    for( int j = 1; j <= m; j++ ){      add(i,j+n,1);    }  }  for( int i = 1; i <= m; i++ ){    int x;scanf("%d",&x);    add(i+n,t,x);  }  int ret = max_flow(s,t);  if(ret < sum){    puts("0");return 0;  }  puts("1");  for( int i = 1; i <= n; i++ ){    for( int j = 0; j < G[i].size(); j++ ){      if(G[i][j].to > n && G[i][j].cap == 0){        printf("%d ",G[i][j].to - n);      }    }    puts("");  }  return 0;}

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