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题目大意:给一张网络,问是否存在一条恰为C的流。若不存在,那是否存在一条弧,使得改动这条弧的容量后能恰有为C的流?
题目分析:先找出最大流,如果最大流不比C小,那么一定存在一条恰为C的流。否则,找出最小割集,然后枚举每一条弧改动其容量,看是否存在恰为C的流。
代码如下:
# include<iostream> # include<cstdio> # include<cmath> # include<string> # include<vector> # include<list> # include<set> # include<map> # include<queue> # include<cstring> # include<algorithm> using namespace std; # define LL long long # define REP(i,s,n) for(int i=s;i<n;++i) # define CL(a,b) memset(a,b,sizeof(a)) # define CLL(a,b,n) fill(a,a+n,b) const double inf=1e30; const int INF=1<<30; const int maxn=10005; struct Edge { int fr,to; LL cap,fw; Edge(int _fr,int _to,LL _cap,LL _fw):fr(_fr),to(_to),cap(_cap),fw(_fw){} bool operator <(const Edge &a) const{ if(fr==a.fr) return to<a.to; return fr<a.fr; } }; vector<Edge>edges,tedges,ans; vector<int>G[2*maxn]; vector<int>minCut; int N,E,cur[105],vis[105],d[105]; LL C,Flow; int s,t; void init() { Flow=0; s=1,t=N; edges.clear(); REP(i,0,N+1) G[i].clear(); } void addEdge(int u,int v,LL c) { edges.push_back(Edge(u,v,c,0)); edges.push_back(Edge(v,u,0,0)); int m=edges.size(); G[u].push_back(m-2); G[v].push_back(m-1); } bool BFS() { CL(vis,0); queue<int>q; vis[s]=1; d[s]=0; q.push(s); while(!q.empty()){ int x=q.front();q.pop(); REP(i,0,G[x].size()){ Edge &e=edges[G[x][i]]; if(!vis[e.to]&&e.cap>e.fw){ vis[e.to]=1; d[e.to]=d[x]+1; q.push(e.to); } } } return vis[t]; } LL DFS(int x,LL a) { if(x==t||a==0) return a; LL flow=0,f; for(int &i=cur[x];i<G[x].size();++i){ Edge &e=edges[G[x][i]]; if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.fw)))>0){ e.fw+=f; edges[G[x][i]^1].fw-=f; flow+=f; a-=f; if(a==0) break; } } return flow; } bool Dinic() { while(BFS()){ CL(cur,0); Flow+=DFS(s,INF); if(Flow>=C) return true; } return false; } void findCut() { minCut.clear(); for(int i=0;i<edges.size();i+=2){ Edge &e=edges[i]; if(vis[e.fr]&&!vis[e.to]) minCut.push_back(i); } } void toFind() { findCut(); ans.clear(); tedges.clear(); int len=edges.size(); REP(i,0,len) tedges.push_back(edges[i]); LL f=Flow; REP(i,0,minCut.size()){ edges[minCut[i]].cap+=C; if(Dinic()) ans.push_back(edges[minCut[i]]); Flow=f; edges.clear(); REP(j,0,len) edges.push_back(tedges[j]); } } void solve(int cas) { printf("Case %d: ",cas); if(C==0||Dinic()){ printf("possible\n"); }else{ toFind(); int len=ans.size(); if(len>0){ printf("possible option:"); sort(ans.begin(),ans.end()); REP(i,0,len) printf("(%d,%d)%c",ans[i].fr,ans[i].to,(i==len-1)?‘\n‘:‘,‘); }else printf("not possible\n"); } } void read() { int a,b; LL c; while(E--) { scanf("%d%d%lld",&a,&b,&c); addEdge(a,b,c); } } int main() { int cas=0; while(scanf("%d%d%lld",&N,&E,&C),N|E|C) { init(); read(); solve(++cas); } return 0; }
UVA-11248 Frequency Hopping (最大流+最小割)
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原文地址:http://www.cnblogs.com/20143605--pcx/p/5027979.html