P2604 ZJOI2010 网络扩容,费用流裸题
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                                P2604 ZJOI2010 网络扩容,费用流裸题
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                                網絡擴容
題目鏈接
https://www.luogu.org/problemnew/show/P2604
題解
對于每條邊u→vu \rightarrow vu→v,我們將按照容量=C=C=C,費用=0=0=0建邊,這些算是免費邊.
 然后我們再對同一對點u,vu,vu,v建立一條容量為INFINFINF,費用為擴容費用WWW的邊,這些邊算是擴容流量要走的邊,有費用.
 跑一邊MCMFMCMFMCMF即可.
代碼
#include <iostream> #include <algorithm> #include <cstring> #include <queue> #define pr(x) std::cout << #x << ':' << x << std::endl #define rep(i,a,b) for(int i = a;i <= b;++i)const int inf = 0x3f3f3f3f; const int mm = 111111; const int maxn = 1111; int node,src,dest,edge; int ver[mm],flow[mm],cst[mm],nxt[mm]; int head[maxn],work[maxn],dis[maxn],q[maxn]; int tot_cost; void prepare(int _node,int _src,int _dest) {node=_node,src=_src,dest=_dest;for(int i=0; i<node; ++i)head[i]=-1;edge=0;tot_cost = 0; } void add_edge(int u,int v,int c,int cost) {ver[edge]=v,flow[edge]=c,nxt[edge]=head[u],cst[edge]=cost,head[u]=edge++;ver[edge]=u,flow[edge]=0,nxt[edge]=head[v],cst[edge]=-cost,head[v]=edge++; } int ins[maxn]; int pre[maxn]; bool Dinic_spfa() {memset(ins,0,sizeof(ins));memset(dis,inf,sizeof(dis));memset(pre,-1,sizeof(pre));std::queue<int> Q;Q.push(src);dis[src] = 0,ins[src] = 1;pre[src] = -1;while(!Q.empty()){int u = Q.front();Q.pop();ins[u] = 0;for(int e = head[u];e != -1;e = nxt[e]){int v = ver[e];if(!flow[e]) continue;if(dis[v] > dis[u] + cst[e]){dis[v] = dis[u] + cst[e];pre[v] = e;if(!ins[v]) ins[v] = 1,Q.push(v);}}}return dis[dest] < inf; } int Dinic_flow() {int i,ret=0,delta=inf;while(Dinic_spfa()){for(int i=pre[dest];i != -1;i = pre[ver[i^1]])delta = std::min(delta,flow[i]);for(int i=pre[dest];i != -1;i = pre[ver[i^1]])flow[i] -= delta,flow[i^1] += delta;ret+=delta;tot_cost += dis[dest]*delta;}return ret; } const int N = 5010; int n,m,k; int u[N],v[N],c[N],w[N]; int main() {std::cin >> n >> m >> k;for(int i = 1;i <= m;++i) {std::cin >> u[i] >> v[i] >> c[i] >> w[i];} prepare(n,0,n-1);for(int i = 1;i <= m;++i) {add_edge(u[i]-1,v[i]-1,c[i],0);}int pf = Dinic_flow();std::cout << pf << " ";prepare(n+1,0,n);for(int i = 1;i <= m;++i) {add_edge(u[i],v[i],c[i],0);add_edge(u[i],v[i],inf,w[i]);}add_edge(0,1,pf+k,0);Dinic_flow();std::cout << tot_cost << std::endl;return 0; }總結
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