Farm Tour
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18961 Accepted: 7326 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 2Sample Output
6
The meaning of the question: there are n points, m edges, undirected graph, what is the shortest path of 1->n->1. FJ took friends to visit, and he might see a few more places, so each side can only be walked once.
Idea: Take the path length as the cost and the flow as 1, so that each path can only be taken once. Build your own source point s, sink point t. s->1 flow 2, n->t flow 2, because these two edges have to go twice. Then the direct expense flow. For the water question, just change the board, the second question after learning the cost flow.
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <queue> 6 7 using namespace std; 8 const int MAXN = 5010; 9 const int MAXM = 50010; 10 const int INF = 0x7FFFFFFF; 11 12 int n, m, first[MAXN], s, t, sign; 13 14 int max_flow, min_cost; 15 16 struct Edge { 17 int to, cap, cost, next; 18 } edge[MAXM * 2]; 19 20 inline void init() { 21 for(int i = 0; i <= n + 1; i++ ) { 22 first[i] = -1; 23 } 24 sign = 0; 25 } 26 27 inline void add_edge(int u, int v, int cap, int cost) { 28 edge[sign].to = v, edge[sign].cap = cap, edge[sign].cost = cost; 29 edge[sign].next = first[u], first[u] = sign ++; 30 edge[sign].to = u, edge[sign].cap = 0, edge[sign].cost = -cost; 31 edge[sign].next = first[v], first[v] = sign ++; 32 } 33 34 int dist[MAXN], inq[MAXN], pre[MAXN], incf[MAXN]; 35 36 bool spfa(int s, int t) { 37 for(int i = 0; i <= n + 1; i++ ) { 38 dist[i] = INF, inq[i] = 0; 39 } 40 queue<int>que; 41 que.push(s), inq[s] = 1, dist[s] = 0; 42 incf[s] = 0x3f3f3f3f; 43 while(!que.empty()) { 44 int now = que.front(); 45 que.pop(); 46 inq[now] = 0; 47 for(int i = first[now]; ~i; i = edge[i].next) { 48 int to = edge[i].to, cap = edge[i].cap, cost = edge[i].cost; 49 if(cap > 0 && dist[to] > dist[now] + cost) { 50 dist[to] = dist[now] + cost; 51 incf[to] = min(incf[now], cap); 52 pre[to] = i; 53 if(!inq[to]) { 54 que.push(to); 55 inq[to] = 1; 56 } 57 } 58 } 59 } 60 return dist[t] != INF; 61 } 62 63 void update(int s, int t) { 64 int x = t; 65 while(x != s) { 66 int pos = pre[x]; 67 edge[pos].cap -= incf[t]; 68 edge[pos ^ 1].cap += incf[t]; 69 x = edge[pos ^ 1].to; 70 } 71 max_flow += incf[t]; 72 min_cost += dist[t] * incf[t]; 73 } 74 75 void minCostMaxFlow(int s, int t) { 76 while(spfa(s, t)) { 77 update(s, t); 78 } 79 } 80 81 int main() 82 { 83 while(~scanf("%d %d", &n, &m)) { 84 s = 0, t = n + 1; 85 init(); 86 for(int i = 1; i <= m; i++ ) { 87 int u, v, cap, cost; 88 scanf("%d %d %d", &u, &v, &cost); 89 add_edge(u, v, 1, cost); 90 add_edge(v, u, 1, cost); 91 } 92 add_edge(s, 1, 2, 0); 93 add_edge(n, t, 2, 0); 94 max_flow = min_cost = 0; 95 minCostMaxFlow(s, t); 96 printf("%d\n", min_cost); 97 } 98 99 return 0; 100 }