TOJ 1583: Farm Tour -- 最小費用最大流 MCMF

1583: Farm Tour

Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte

Total Submit: 12            Accepted:5

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 Feburary 2003

分析：最小費用最大流模闆題。

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                      #define mp make_pair using namespace std; typedef unsigned int ui; typedef long long ll; typedef unsigned long long ull; typedef pair
                     
                       pii; typedef vector
                      
                        vi; typedef vi::iterator vi_it; typedef map
                       
                         mii; typedef priority_queue
                        
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                            > rpqi; const int MAX_NODE = 1000 + 10; const int MAX_EDGE = 10000 + 10; const int INF = (int)1.0e9; int p[MAX_NODE]; //predecessor array int d[MAX_NODE]; //cost array(from the source) bool inq[MAX_NODE]; //in-queue flag int cnt = 0; //edge counting int head[MAX_NODE]; struct { int u; int v; int cap; int cost; int next; } edge[MAX_EDGE << 2]; void add_edge(int u, int v, int cap, int cost) { edge[cnt].u = u; edge[cnt].v = v; edge[cnt].cap = cap; edge[cnt].cost = cost; edge[cnt].next = head[u]; head[u] = cnt++; edge[cnt].u = v; edge[cnt].v = u; edge[cnt].cap = 0; edge[cnt].cost = -cost; edge[cnt].next = head[v]; head[v] = cnt++; } class MCMF { public: /** Description: Arguments: n -- the number of nodes s -- the source t -- the destination Returns:pair
                           
                             */ pii mcmf(int n, int s, int t); }; pii MCMF::mcmf(int n, int s, int t) { queue
                            
                              q; int c = 0; //min cost int f = 0; //max flow int u, v, i; while (true) { memset(inq, 0, sizeof(inq)); for (i = 0; i < n; ++i) { d[i] = INF; } d[s] = 0; q.push(s); inq[s] = true; p[s] = -1; while (!q.empty()) { u = q.front(); q.pop(); inq[u] = false; for (i = head[u]; i != -1; i = edge[i].next) { v = edge[i].v; if (edge[i].cap > 0 && d[v] > d[u] + edge[i].cost) { d[v] = d[u] + edge[i].cost; p[v] = i; if (!inq[v]) { q.push(v); inq[v] = true; } } } } if (d[t] == INF) { return mp(f, c); } int a = INF; for (i = p[t]; i != -1; i = p[edge[i].u]) { a = min(a, edge[i].cap); } for (i = p[t]; i != -1; i = p[edge[i].u]) { edge[i].cap -= a; edge[i ^ 1].cap += a; } c += d[t] * a; f += a; } return mp(-1, -1); //an error occurs } int main(int argc, char *argv[]) { // freopen("D:\\in.txt", "r", stdin); int n, m; cin >> n >> m; memset(head, -1, sizeof(head)); while (m--) { int u, v, w; scanf("%d%d%d", &u, &v, &w); add_edge(u, v, 1, w); add_edge(v, u, 1, w); } add_edge(0, 1, 2, 0); add_edge(n, n + 1, 2, 0); MCMF mc; pii ans = mc.mcmf(n + 2, 0, n + 1); assert(ans.first == 2); cout << ans.second << endl; return 0; }