PAT-Top-1003 Universal Travel Sites (35分)網絡流最大流

1003 Universal Travel Sites (35分)

題目傳送門：1003 Universal Travel Sites (35分)

一、題目大意

二、解題思路

網絡流問題，第一次嘗試，看了劉汝佳的《算法競賽入門經典》上的例子和代碼，理解了後自己寫了出來，居然真的AC了，😆😝☺️

三、AC代碼

#include<bits/stdc++.h>
using namespace std;
const int N = 5000;
struct Edge{
  int from, to, cap, flow;
  Edge(int from, int to, int cap, int flow):from(from), to(to), cap(cap), flow(flow){}
};
map<string, int>Map;
vector<Edge>edges;
vector<int>G[N];
void addEdge(int from, int to, int cap, int flow){
  edges.push_back(Edge(from, to, cap, 0));// 正向邊
  edges.push_back(Edge(to, from, 0, 0)); // 反向邊
  int cnt = edges.size();
  G[from].push_back(cnt-2);// 記住每條鄰邊的編號
  G[to].push_back(cnt-1);
}
int main(){
  freopen("input.txt", "r", stdin);
  int no = 1, startNo = 1, endNo = 2;
  string stra, strb;
  cin >> stra >> strb;
  Map[stra] = no++;
  Map[strb] = no++;
  int n;
  cin >> n;
  int x;
  for(int i = 0; i < n; i++){
    cin >> stra >> strb >> x;
    if(!Map[stra])
      Map[stra] = no++;
    if(!Map[strb])
      Map[strb] = no++;
    addEdge(Map[stra], Map[strb], x, 0);
  }
  int flow = 0;
  for(;;){
    int a[no], father[no];
    memset(a, 0, sizeof(a));
    memset(father, 0, sizeof(father));
    a[startNo] = INT32_MAX;
    queue<int>Q;
    Q.push(startNo);
    while(!Q.empty()){
      int head = Q.front();
      Q.pop();
      for(int i = 0; i < G[head].size(); i++){
        Edge& e = edges[G[head][i]];
        if(!a[e.to] and e.flow < e.cap){
          a[e.to] = min(a[head], e.cap - e.flow);
          father[e.to] = G[head][i];
          Q.push(e.to);
        }
      }
      if(a[endNo]){// 終點
        break;// 完成一次增廣路徑
      }
    }
    if(!a[endNo]){
      break;// 無法再增廣，退出循環
    }
    for(int u = endNo; u != startNo; u = edges[father[u]].from){
      edges[father[u]].flow += a[endNo];
      edges[father[u]^1].flow -= a[endNo];
    }
    flow += a[endNo];
  }
  cout << flow << endl;
}