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.
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.
* 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.
题目大意：给一个n个点m条边的无向图，求从1走到n再从n走到1的最短路径，要求边不重复。
思路：建立一个附加源点S，连一条边到点1，容量为2，费用为0；建立一个附加汇点T，从n连一条边到T，容量为2，费用为0；对于每条边u、v，连一条边u到v，容量为1，费用为距离，再连一条边v到u，容量为1，费用为距离。最小费用最大流为答案。
建图正确性略微说明：可以考虑两次从1走到n，边不重复，那么建图流两个流量从源点到汇点，最小的费用就保证了距离最短。由于边长度为非负数，那么最小花费一定不会是一条路径从u到v，另一条路径从v到u，如果这样走了两次，还不如两条路径都不经过u-v，都走另外一条路的后继路径（不要跟我说长度是0怎么破）
PS：忘了改数组大小又TLE了一次T_T，怎么老是忘
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 const int MAXV = 1010; 9 const int MAXE = 40010; 10 const int INF = 0x7f7f7f7f; 11 12 struct SPFA_COST_FLOW { 13 bool vis[MAXV]; 14 int head[MAXV], dis[MAXV], pre[MAXV]; 15 int to[MAXE], next[MAXE], cost[MAXE], flow[MAXE]; 16 int n, st, ed, ecnt; 17 18 void init() { 19 memset(head, 0, sizeof(head)); 20 ecnt = 2; 21 } 22 23 void add_edge(int u, int v, int c, int w) { 24 to[ecnt] = v; flow[ecnt] = c; cost[ecnt] = w; next[ecnt] = head[u]; head[u] = ecnt++; 25 to[ecnt] = u; flow[ecnt] = 0; cost[ecnt] = -w; next[ecnt] = head[v]; head[v] = ecnt++; 26 } 27 28 bool spfa() { 29 memset(vis, 0, sizeof(vis)); 30 memset(dis, 0x7f, sizeof(dis)); 31 queue<int> que; que.push(st); 32 vis[st] = true; dis[st] = 0; 33 while(!que.empty()) { 34 int u = que.front(); que.pop(); 35 vis[u] = false; 36 for(int p = head[u]; p; p = next[p]) { 37 int &v = to[p]; 38 if(flow[p] && dis[v] > dis[u] + cost[p]) { 39 dis[v] = dis[u] + cost[p]; 40 pre[v] = p; 41 if(!vis[v]) { 42 vis[v] = true; 43 que.push(v); 44 } 45 } 46 } 47 } 48 return dis[ed] < INF; 49 } 50 51 int maxFlow, minCost; 52 53 int min_cost_flow(int ss, int tt, int nn) { 54 st = ss, ed = tt, n = nn; 55 maxFlow = minCost = 0; 56 while(spfa()) { 57 minCost += dis[ed]; 58 int u = ed, tmp = INF; 59 while(u != st) { 60 tmp = min(tmp, flow[pre[u]]); 61 u = to[pre[u] ^ 1]; 62 } 63 u = ed; 64 while(u != st) { 65 flow[pre[u]] -= tmp; 66 flow[pre[u] ^ 1] += tmp; 67 u = to[pre[u] ^ 1]; 68 } 69 maxFlow += tmp; 70 } 71 return minCost; 72 } 73 } G; 74 75 int n, m; 76 77 int main() { 78 scanf("%d%d", &n, &m); 79 G.init(); 80 int ss = n + 1, tt = n + 2; 81 while(m--) { 82 int u, v, c; 83 scanf("%d%d%d", &u, &v, &c); 84 G.add_edge(u, v, 1, c); 85 G.add_edge(v, u, 1, c); 86 } 87 G.add_edge(ss, 1, 2, 0); 88 G.add_edge(n, tt, 2, 0); 89 printf("%d ", G.min_cost_flow(ss, tt, tt)); 90 }