UVA-11380 - Down Went The Titanic(最大流-拆点)
题意比较复杂就不多说了...
分析:只是拆点即可,对于@说不能同时有两个人站在上面,可以直接忽略即可,因为这个是不沉没的,故一个一个通过即可.
// File Name: 11380.cpp // Author: Zlbing // Created Time: 2013/4/26 15:12:42 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack> #include<cmath> #include<queue> using namespace std; #define CL(x,v); memset(x,v,sizeof(x)); #define INF 0x3f3f3f3f #define LL long long #define REP(i,r,n) for(int i=r;i<=n;i++) #define RREP(i,n,r) for(int i=n;i>=r;i--) const int MAXN=2e3+100; struct Edge{ int from,to,cap,flow; }; bool cmp(const Edge& a,const Edge& b){ return a.from < b.from || (a.from == b.from && a.to < b.to); } struct Dinic{ int n,m,s,t; vector<Edge> edges; vector<int> G[MAXN]; bool vis[MAXN]; int d[MAXN]; int cur[MAXN]; void init(int n){ this->n=n; for(int i=0;i<=n;i++)G[i].clear(); edges.clear(); } void AddEdge(int from,int to,int cap){ edges.push_back((Edge){from,to,cap,0}); edges.push_back((Edge){to,from,0,0});//当是无向图时,反向边容量也是cap,有向边时,反向边容量是0 m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BFS(){ CL(vis,0); queue<int> Q; Q.push(s); d[s]=0; vis[s]=1; while(!Q.empty()){ int x=Q.front(); Q.pop(); for(int i=0;i<G[x].size();i++){ Edge& e=edges[G[x][i]]; if(!vis[e.to]&&e.cap>e.flow){ vis[e.to]=1; d[e.to]=d[x]+1; Q.push(e.to); } } } return vis[t]; } int DFS(int x,int a){ if(x==t||a==0)return a; int 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.flow)))>0){ e.flow+=f; edges[G[x][i]^1].flow-=f; flow+=f; a-=f; if(a==0)break; } } return flow; } //当所求流量大于need时就退出,降低时间 int Maxflow(int s,int t,int need){ this->s=s;this->t=t; int flow=0; while(BFS()){ CL(cur,0); flow+=DFS(s,INF); if(flow>need)return flow; } return flow; } //最小割割边 vector<int> Mincut(){ BFS(); vector<int> ans; for(int i=0;i<edges.size();i++){ Edge& e=edges[i]; if(vis[e.from]&&!vis[e.to]&&e.cap>0)ans.push_back(i); } return ans; } void Reduce(){ for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow; } void ClearFlow(){ for(int i = 0; i < edges.size(); i++) edges[i].flow = 0; } }; int n,m,p; char ch[100][100]; Dinic solver; int dx[4]={-1,1,0,0}; int dy[4]={0,0,1,-1}; int main() { while(~scanf("%d%d%d",&n,&m,&p)) { REP(i,0,n-1) scanf("%s",ch[i]); solver.init(n*m*2+1); int s=0,t=1; REP(i,0,n-1) { REP(j,0,m-1) { if(ch[i][j]=='~')continue; REP(k,0,3) { int nx=i+dx[k]; int ny=j+dy[k]; if(nx<0||nx>=n||ny<0||ny>=m)continue; int a=(i*m+j+1)*2+1; int b=(nx*m+ny+1)*2; solver.AddEdge(a,b,INF); } int a=(i*m+j+1)*2; int b=(i*m+j+1)*2+1; if(ch[i][j]=='.') { solver.AddEdge(a,b,1); } else if(ch[i][j]=='*') { solver.AddEdge(s,a,1); solver.AddEdge(a,b,1); } else if(ch[i][j]=='@') { solver.AddEdge(a,b,INF); } else if(ch[i][j]=='#') { solver.AddEdge(a,b,INF); solver.AddEdge(b,t,p); } } } int ans=solver.Maxflow(s,t,INF); printf("%d\n",ans); } return 0; }