Uva 10806 来回最短路，不重复，MCMF

Posted 2020-09-04 树的种子

题目链接：https://uva.onlinejudge.org/external/108/10806.pdf

题意：无向图，从1到n来回的最短路，不走重复路。

分析：可以考虑为1到n的流量为2时的最小花费；

建图： 一个点到一个点的容量为1，费用为距离。

  1 #include <cstring>
  2 #include <cstdio>
  3 #include <vector>
  4 #include <queue>
  5 #include <algorithm>
  6 #include <iostream>
  7 using namespace std;
  8 
  9 const int maxn = 100 + 10, INF = 1000000000;
 10 
 11 
 12 struct Edge
 13 {
 14     int from, to, cap, flow, cost;
 15 };
 16 
 17 struct MCMF
 18 {
 19     int n, m;
 20     vector<Edge> edges;
 21     vector<int> G[maxn];
 22     bool inq[maxn];         // 是否在队列中
 23     int d[maxn];           // Bellman-Ford
 24     int p[maxn];           // 上一条弧
 25     int a[maxn];           // 可改进量
 26 
 27     void init(int n)
 28     {
 29         this->n = n;
 30         for(int i = 0; i < n; i++) G[i].clear();
 31         edges.clear();
 32     }
 33 
 34     void AddEdge(int from, int to, int cap, int cost)
 35     {
 36         edges.push_back((Edge)
 37         {
 38             from, to, cap, 0, cost
 39         });
 40         edges.push_back((Edge)
 41         {
 42             to, from, 0, 0, -cost
 43         });
 44         m = edges.size();
 45         G[from].push_back(m-2);
 46         G[to].push_back(m-1);
 47     }
 48 
 49     bool BellmanFord(int s, int t, int &flow, long long& cost)
 50     {
 51         memset(inq,0,sizeof(inq));
 52         for(int i=0;i<n;i++)
 53             d[i] = INF;
 54         d[s] = 0;
 55         inq[s] = true;
 56         p[s] = 0;
 57         a[s] = INF;
 58 
 59         queue<int> Q;
 60         Q.push(s);
 61         while(!Q.empty())
 62         {
 63             int u = Q.front();
 64             Q.pop();
 65             inq[u] = false;
 66             for(int i = 0; i < G[u].size(); i++)
 67             {
 68                 Edge& e = edges[G[u][i]];
 69                 if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
 70                 {
 71                     d[e.to] = d[u] + e.cost;
 72                     p[e.to] = G[u][i];
 73                     a[e.to] = min(a[u], e.cap - e.flow);
 74                     if(!inq[e.to])
 75                     {
 76                         Q.push(e.to);
 77                         inq[e.to] = true;
 78                     }
 79                 }
 80             }
 81         }
 82         if(d[t] == INF) return false; //s-t 不连通，失败退出
 83         flow += a[t];
 84         cost += (long long)d[t] * (long long)a[t];
 85         int u = t;
 86         while(u != s)
 87         {
 88             edges[p[u]].flow += a[t];
 89             edges[p[u]^1].flow -= a[t];
 90             u = edges[p[u]].from;
 91         }
 92         return true;
 93     }
 94 
 95      pair<long long,int>Mincost(int s, int t)
 96     {
 97         long long cost = 0;
 98         int flow = 0;
 99         while(BellmanFord(s, t, flow, cost));
100         return pair<long long ,int>{cost,flow};
101     }
102 };
103 
104 MCMF solver;
105 int n;
106 
107 
108 int main()
109 {
110     while(true)
111     {
112         scanf("%d", &n);
113         if(n == 0) break;
114         solver.init(n + 1);
115         int m, from, to, cost;
116         scanf("%d", &m);
117         for(int i = 0; i < m; i++)
118         {
119             scanf("%d%d%d", &from, &to, &cost);
120             solver.AddEdge(from, to, 1, cost);
121             solver.AddEdge(to, from, 1, cost);
122         }
123         solver.AddEdge(0, 1, 2, 0);
124 
125         pair<long long,int> ans = solver.Mincost(0, n);
126         int flow = ans.second;
127 
128         if(flow != 2)
129             puts("Back to jail");
130         else
131             printf("%lld\n", ans.first);
132     }
133     return 0;
134 }