Drainage Ditches |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 45 Accepted Submission(s): 38 |
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Problem Description
Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle. |
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Input
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
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Output
For each case, output a single integer, the maximum rate at which water may emptied from the pond. |
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Sample Input
5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10 |
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Sample Output
50 |
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Source
USACO 93
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题意：
裸的最大流
代码：
//Edmonds-Karp算法，紫书366页。模板。点的编号从0开始。 #include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> using namespace std; const int maxn=202,inf=0x7fffffff; struct edge{ int from,to,cap,flow; edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){} }; struct Edmonds_Karp{ int n,m; vector<edge>edges;//边数的两倍 vector<int>g[maxn];//邻接表，g[i][j]表示节点i的第j条边在e数组中的序号 int a[maxn];//当起点到i的可改进量 int p[maxn];//最短路树上p的入弧编号 void init(int 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));//反向弧 m=edges.size(); g[from].push_back(m-2); g[to].push_back(m-1); } int Maxflow(int s,int t){ int flow=0; for(;;){ memset(a,0,sizeof(a)); queue<int>q; q.push(s); a[s]=inf; while(!q.empty()){ int x=q.front();q.pop(); for(int i=0;i<(int)g[x].size();i++){ edge&e=edges[g[x][i]]; if(!a[e.to]&&e.cap>e.flow){ p[e.to]=g[x][i]; a[e.to]=min(a[x],e.cap-e.flow); q.push(e.to); } } if(a[t]) break; } if(!a[t]) break; for(int u=t;u!=s;u=edges[p[u]].from){ edges[p[u]].flow+=a[t]; edges[p[u]^1].flow-=a[t]; } flow+=a[t]; } return flow; } }EK; int main() { int n,m,a,b,c; while(scanf("%d%d",&n,&m)==2){ EK.init(m); for(int i=0;i<n;i++){ scanf("%d%d%d",&a,&b,&c); a--;b--; EK.addedge(a,b,c); } printf("%d ",EK.Maxflow(0,m-1)); } return 0; }
//Dinic算法模板 白书358页，点的编号从0开始 #include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> using namespace std; const int maxn=202; const int inf=0x7fffffff; struct Edge{ int from,to,cap,flow; Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){} }; 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));//反向弧 m=edges.size(); g[from].push_back(m-2); g[to].push_back(m-1); } bool Bfs(){ memset(vis,0,sizeof(vis)); 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<(int)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<(int)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; } int Maxflow(int s,int t){ this->s=s;this->t=t; int flow=0; while(Bfs()){ memset(cur,0,sizeof(cur)); flow+=Dfs(s,inf); } return flow; } }dc; int main() { int n,m,a,b,c; while(scanf("%d%d",&n,&m)==2){ dc.Init(m); while(n--){ scanf("%d%d%d",&a,&b,&c); a--;b--; dc.Addedge(a,b,c); } printf("%d ",dc.Maxflow(0,m-1)); } return 0; }
//anather #include<iostream> #include<cstring> #include<cstdio> using namespace std; const int INF=0x7fffffff; const int MAXN=10002;//点数 const int MAXM=10002;//边数 int n,m,tot,S,T,head[MAXN],h[MAXN],q[MAXN],ans; struct Edge { int to,val,next; }edge[MAXM]; void init(int last) { S=0;T=last-1;//S源点，T汇点 tot=0; memset(head,-1,sizeof(head)); } void addedge(int x,int y,int z) { edge[tot].to=y;edge[tot].val=z;edge[tot].next=head[x]; head[x]=tot++; } bool bfs() { memset(h,-1,sizeof(h)); int top=0,last=1; q[top]=S;h[S]=0; while(top<last){ int now=q[top];top++; for(int i=head[now];i!=-1;i=edge[i].next){ if(edge[i].val&&h[edge[i].to]<0){ q[last++]=edge[i].to; h[edge[i].to]=h[now]+1; } } } if(h[T]==-1) return 0; return 1; } int dfs(int x,int f) { if(x==T) return f; int w,used=0; for(int i=head[x];i!=-1;i=edge[i].next){ if(edge[i].val&&h[edge[i].to]==h[x]+1){ w=f-used; w=dfs(edge[i].to,min(w,edge[i].val)); edge[i].val-=w; edge[i^1].val+=w; used+=w; if(used==f) return f; } } if(!used) h[x]=-1; return used; } int dinic() { int ans=0; while(bfs()) ans+=dfs(S,INF); return ans; } int main() { int n,m,a,b,c; while(scanf("%d%d",&n,&m)==2){ init(m);//根据题目传参 while(n--){ scanf("%d%d%d",&a,&b,&c); a--;b--; addedge(a,b,c); addedge(b,a,0);//建反向边 //建边根据题目而定 } int ans=dinic(); printf("%d ",ans); } return 0; }