Admiral UVA

将题目抽象成为最小费用网络流的问题，其中对于每个中间节点i都分成两个节点i和j，同时建立一条从i到j的边，这条边的容量为1，花费为0，同时对于输入的每条边，都建立对应的边，边的容量也为1，同时花费为相应的权值，那么问题就转化为求出从开始点到终点的流量为2的最小花费，按照紫书的算法，结合BellmanFloyd即可求出结果，但是要注意一点我们要将流量的上限设置为2，在具体实现的时候注意细节方面的改动即可，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;class Edge{public:int from, to, cap, flow, cost;Edge(int fr, int t, int ca, int fl, int c):from(fr),to(t),cap(ca),flow(fl),cost(c){}};class Solve{public:int v, e;vector<Edge> edge;vector<int> G[2005];int flow_limit = 2;const int Inf = 0x3f3f3f3f;void addEdge(int from,int to,int cap,int cost){edge.push_back(Edge(from, to, cap, 0, cost));edge.push_back(Edge(to, from, 0, 0, -cost));int m = edge.size() - 1;G[from].push_back(m - 1);G[to].push_back(m);}bool BellmanFord(int start,int end,int& flow,int& cost){int inq[2005];int parent[2005];int price[2005];int Flow[2005];Flow[0] = Inf;memset(price, Inf, sizeof(price));price[0] = 0;memset(inq, 0, sizeof(inq));inq[start] = 1;queue<int> q;q.push(start);while (!q.empty()){int id = q.front();q.pop();inq[id] = 0;for (int i = 0; i < G[id].size(); i++){int ide = G[id][i];int to = edge[ide].to;if (edge[ide].cap > edge[ide].flow&&price[to] > price[id] + edge[ide].cost){Flow[to] = min(Flow[id], edge[ide].cap - edge[ide].flow);price[to] = price[id] + edge[ide].cost;parent[to] = ide;if (!inq[to]){q.push(to);inq[to] = 1;}}}}if (price[end] == Inf) return false;if (flow + Flow[end] > flow_limit) Flow[end] = flow_limit - flow;cost += Flow[end] * price[end];flow += Flow[end];for (int i = end; i != start; i = edge[parent[i]].from){edge[parent[i]].flow += Flow[end];edge[parent[i] ^ 1].flow -= Flow[end];}return true;}void MaxFlow(int start,int end){int cost = 0, flow=0;while (flow < flow_limit&&BellmanFord(start, end, flow, cost));cout << cost << endl;}void Init(){edge.clear();for (int i = 0; i < 2005; i++) G[i].clear();for (int i = 1; i <= v - 2; i++){int id2 = i + v - 1;addEdge(i, id2, 1, 0);}for (int i = 0; i < e; i++){int a, b, c;cin >> a >> b >> c;a--;if (a != 0 && a != v - 1) a += v - 1;b--;addEdge(a, b, 1, c);}}void Deal(){Init();MaxFlow(0,v-1);}};int main(){Solve a;while (cin >> a.v >> a.e){a.Deal();}return 0;}