Farm Tour(最小費用最大流模板)
阿新 • • 發佈：2017-11-07
arc cost -c fields ros lap view which accept
To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.
He wants his tour to be as short as possible, however he doesn‘t want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.
* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path‘s length.
Farm Tour
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18150 | Accepted: 7023 |
Description
When FJ‘s friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.
He wants his tour to be as short as possible, however he doesn‘t want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.
Input
* Line 1: Two space-separated integers: N and M.* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path‘s length.
Output
A single line containing the length of the shortest tour.Sample Input
4 5 1 2 1 2 3 1 3 4 1 1 3 2 2 4 2
Sample Output
6
Source
USACO 2003 February Green //題意：n 個點， m 條無向邊，要求從 1 走到 n 點，再從 n 走到 1 點，不能走重復路，求最短路徑 //可以看成從 1 走到 n 用兩種路徑，因為只能走一遍，所以邊的容量為 1 所以建立一個附加的源點容量為2，費用為 0 到1點，附加的匯點容量為 2 ，費用為 0 到 n+1 點跑最小費用最大流即可
1 //# include <bits/stdc++.h> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <queue> 6 using namespace std; 7 # define eps 1e-8 8 # define INF 0x3f3f3f3f 9 # define pi acos(-1.0) 10 # define MXN 1005 11 # define MXM 20050 12 13 struct Edge{ 14 int from, to, flow, cost, cap; 15 }edges[MXM*2]; //有向邊數*2 16 17 struct MCMF{ 18 int n, m, idx; //點，邊，邊數 19 int flow, cost; 20 vector<int> G[MXN]; //記錄邊 21 int inq[MXN]; //BFS用 22 int dis[MXN]; //層次 23 int pre[MXN]; //上一條弧 24 int adf[MXN]; //增量 25 26 void Init(){ 27 idx=0; 28 for (int i=0;i<=n+1;i++) G[i].clear(); //有附加點時要註意 29 } 30 31 void Addedge(int u,int v,int cost,int cap){ 32 edges[idx++] = (Edge){u,v,0,cost,cap}; 33 edges[idx++] = (Edge){v,u,0,-cost,0}; 34 G[u].push_back(idx-2); 35 G[v].push_back(idx-1); 36 } 37 38 int Bellman(int s, int t) 39 { 40 memset(dis,0x3f,sizeof(dis)); //有附加點時要註意 41 memset(inq,0,sizeof(inq)); 42 dis[s]=0, inq[s]=1, adf[s]=INF; 43 queue<int> Q; 44 Q.push(s); 45 while (!Q.empty()) 46 { 47 int u = Q.front(); Q.pop(); 48 inq[u]=0; 49 for (int i=0;i<(int)G[u].size();i++) 50 { 51 Edge &e = edges[G[u][i]]; 52 if (dis[e.to] > dis[u] + e.cost && e.cap > e.flow) 53 { 54 dis[e.to] = dis[u] + e.cost; 55 adf[e.to] = min(adf[u], e.cap-e.flow); 56 pre[e.to] = G[u][i]; 57 if (!inq[e.to]) 58 { 59 Q.push(e.to); 60 inq[e.to]=1; 61 } 62 } 63 } 64 } 65 if (dis[t]==INF) return false; 66 67 flow+=adf[t]; 68 cost+=adf[t]*dis[t]; 69 int x=t; 70 while(x!=s) 71 { 72 edges[pre[x]].flow+=adf[t]; 73 edges[pre[x]^1].flow-=adf[t]; 74 x=edges[pre[x]].from; 75 } 76 return true; 77 } 78 79 int MinCost(int s,int t) 80 { 81 flow = 0, cost = 0; 82 while(Bellman(s, t)); 83 return cost; 84 } 85 }F; 86 87 int main() 88 { 89 scanf("%d%d",&F.n,&F.m); 90 F.Init(); 91 for (int i=1;i<=F.m;i++) 92 { 93 int u,v,cost; 94 scanf("%d%d%d",&u,&v,&cost); 95 F.Addedge(u,v,cost,1); 96 F.Addedge(v,u,cost,1); 97 } 98 F.Addedge(0, 1, 0, 2); 99 F.Addedge(F.n, F.n+1, 0, 2); 100 printf("%d\n",F.MinCost(0,F.n+1)); 101 return 0; 102 }View Code
Farm Tour(最小費用最大流模板)