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POJ 2135 Farm Tour 最小费用流
两条路不能有重边,既每条边的容量是1。求流量为2的最小费用即可。
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("data.in","r",stdin); // freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=‘ ‘&&ch!=‘\n‘)return ch; } return EOF; } struct Edge { int from,to,cost,cap; }; const int maxn = 3111; vector<int> g[maxn]; vector<Edge> edge; int n,m,s,t; void init() { for(int i = 1; i <= n; i++) g[i].clear(); edge.clear(); } void add(int from, int to, int cost, int cap) { edge.push_back((Edge){from, to, cost, cap}); g[from].push_back(edge.size() - 1); edge.push_back((Edge){to, from, -cost, 0}); g[to].push_back(edge.size() - 1); } int d[maxn]; int inq[maxn]; int road[maxn]; int SPFA() { queue<int> q; q.push(s); memset(d, INF, sizeof(d)); memset(inq, 0, sizeof(inq)); inq[s] = true; d[s] = 0; road[s] = -1; while(!q.empty()) { int x = q.front(); q.pop(); inq[x] = false; for(int i = 0; i < g[x].size(); i++) { Edge &e = edge[g[x][i]]; if(e.cap>0 && d[x] + e.cost < d[e.to]) { d[e.to] = d[x] + e.cost; road[e.to] = g[x][i]; if(!inq[e.to]) { inq[e.to] = true; q.push(e.to); } } } } return d[t]; } int max_cost_flow() { int flow = 2; int cost = 0; while(flow) { int d = SPFA(); int f = flow; for(int e = road[t]; e != -1; e = road[edge[e].from]) { Edge &E = edge[e]; f = min(f, E.cap); } flow -= f; cost += d * f; for(int e = road[t]; e != -1; e = road[edge[e].from]) { edge[e].cap -= f; edge[e^1].cap += f; } } return cost; } int main() { debug(); while(scanf("%d%d", &n, &m) != EOF) { init(); for(int i = 1; i <= m; i++) { int from,to,cost; scanf("%d%d%d", &from, &to, &cost); add(from, to, cost, 1); add(to, from, cost, 1); } s=1; t=n; printf("%d\n", max_cost_flow()); } return 0; }
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