POJ 1459 Power Network （Dinic非递归 | SAP）
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Power Network
Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.
 An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6. Input There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.
Output For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.
Sample Input 2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
(3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
(0)5 (1)2 (3)2 (4)1 (5)4
Sample Output 15 6 Hint The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.
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题目大意:
给几个发电站，给几个消耗站，再给几个转发点。
发电站只发电，消耗站只消耗电，转发点只是转发电，再给各个传送线的传电能力。
问你消耗站能获得的最多电是多少
思路：建立一个超级源点和超级汇点，把发电站相连起来，消耗站连起来 然后就是模版的力量了
在此也讲了下dinic的原理：
求最大流的本质，就是不停的寻找增广路径。直到找不到增广路径为止。
对于这个一般性的过程，Dinic算法的优化如下：
（1）
Dinic算法首先对图进行一次BFS，然后在BFS生成的层次图中进行多次DFS。
层次图的意思就是，只有在BFS树中深度相差1的节点才是连接的。
这就切断了原有的图中的许多不必要的连接。很牛逼！
这是需要证明的，估计证明也很复杂。
（2）
除此之外，每次DFS完后，会找到路径中容量最小的一条边。
在这条边之前的路径的容量是大于等于这条边的容量的。
那么从这条边之前的点，可能引发出别的增广路径。
比如说 S -> b -> c -> d -> T 是一条增广路径，容量最小的边是 b -> c。
可能存在一条 S -> b -> e -> f -> g -> T 这样的增广路径。
这样的话，在找到第一条增广路径后，只需要回溯到 b 点，就可以继续找下去了。
这样做的好处是，避免了找到一条路径就从头开始寻找另外一条的开销。
也就是再次从 S 寻找到 b 的开销。
这个过程看似复杂，但是代码实现起来很优雅，因为它的本质就是回溯！
（3）
在同一次 DFS 中。如果从一个点引发不出任何的增广路径，就将这个点在层次图中抹去
#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; const int VM=150; const int EM=20500; const int INF=0x3f3f3f3f; struct Edge{ int u,v,nxt; int cap; }edge[EM]; int n,np,nc,m,cnt,head[VM]; int src,des,dep[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].u=cu; edge[cnt].v=cv; edge[cnt].cap=cw; edge[cnt].nxt=head[cu]; head[cu]=cnt++; edge[cnt].v=cv; edge[cnt].v=cu; edge[cnt].cap=0; edge[cnt].nxt=head[cv]; head[cv]=cnt++; } int BFS(){ queue<int> q; while(!q.empty()) q.pop(); memset(dep,-1,sizeof(dep)); dep[src]=0; q.push(src); while(!q.empty()){ int u=q.front(); q.pop(); for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].v; if(edge[i].cap>0 && dep[v]==-1){ dep[v]=dep[u]+1; q.push(v); } } } if(dep[des]==-1) return 0; return 1; } int Dinic(){ int ans=0; int stack[VM],next[VM]; ////stack[i为栈，存储当前增广路, next[i]存储当前点的后继 跟head是一样的 while(BFS()){ memcpy(next,head,sizeof(head)); int u=src, top=0; //u为当前结点 while(1){ if(u==des){ //增广路已全部进栈 int minx=INF, loc; for(int i=0;i<top;i++) //找最小的增广跟并loc记录其在stack中位置 if(minx>edge[stack[i]].cap){ //以便退回该边继续DFS minx=edge[stack[i]].cap; loc=i; } for(int i=0;i<top;i++){ //将增广路中的所有边修改 edge[stack[i]].cap-=minx; edge[stack[i]^1].cap+=minx; } ans+=minx; top=loc; u=edge[stack[top]].u; //当前结点修改为最小边的起点 } for(int i=next[u];i!=-1;next[u]=i=edge[i].nxt) //找到当前结点对应的下一条边 if(edge[i].cap>0 && dep[edge[i].v]==dep[u]+1) //不满足条件时，修改cur值 break; //（去掉不合适的占）eg：1-->2 1-->3 1-->4 有边 但只有1-->4 这条边满足条件 就把1到2、3的边给去掉 if(next[u]!=-1){ //当前结点的下一条边存在 stack[top++]=next[u]; //把该边放入栈中 u=edge[next[u]].v; //再从下个点开始找 }else{ if(top==0) //当前结点无未遍历的下一条边且栈空，DFS找不到下一条增广路 break; dep[u]=-1; //当前结点不在增广路中，剔除该点 u=edge[stack[--top]].u; //退栈 回朔，继续查找 } } } return ans; } int main(){ //freopen("input.txt","r",stdin); char str[30]; while(~scanf("%d%d%d%d",&n,&np,&nc,&m)){ cnt=0; memset(head,-1,sizeof(head)); src=n, des=n+1; int u,v,w; while(m--){ scanf("%s",str); sscanf(str,"(%d,%d)%d",&u,&v,&w); addedge(u,v,w); } while(np--){ scanf("%s",str); sscanf(str,"(%d)%d",&v,&w); addedge(src,v,w); } while(nc--){ scanf("%s",str); sscanf(str,"(%d)%d",&u,&w); addedge(u,des,w); } printf("%d\n",Dinic()); } return 0; }
递归Dinic：
#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; const int VM=150; const int EM=20500; const int INF=0x3f3f3f3f; struct Edge{ int u,v,nxt; int cap; }edge[EM]; int n,np,nc,m,cnt,head[VM]; int src,des,dep[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].u=cu; edge[cnt].v=cv; edge[cnt].cap=cw; edge[cnt].nxt=head[cu]; head[cu]=cnt++; edge[cnt].v=cv; edge[cnt].v=cu; edge[cnt].cap=0; edge[cnt].nxt=head[cv]; head[cv]=cnt++; } int BFS(){ queue<int> q; while(!q.empty()) q.pop(); memset(dep,-1,sizeof(dep)); dep[src]=0; q.push(src); while(!q.empty()){ int u=q.front(); q.pop(); for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].v; if(edge[i].cap>0 && dep[v]==-1){ dep[v]=dep[u]+1; q.push(v); } } } if(dep[des]==-1) return 0; return 1; } int DFS(int u,int minx){ if(u==des) return minx; int tmp; for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].v; if(edge[i].cap>0 && dep[v]==dep[u]+1 && (tmp=DFS(v,min(minx,edge[i].cap)))){ edge[i].cap-=tmp; edge[i^1].cap+=tmp; return tmp; } } dep[u]=-1; //加了这句，，递归也AC return 0; } int Dinic(){ int ans=0,tmp; while(BFS()){ while(1){ tmp=DFS(src,INF); if(tmp==0) break; ans+=tmp; } } return ans; } int main(){ //freopen("input.txt","r",stdin); char str[30]; while(~scanf("%d%d%d%d",&n,&np,&nc,&m)){ cnt=0; memset(head,-1,sizeof(head)); src=n, des=n+1; int u,v,w; while(m--){ scanf("%s",str); sscanf(str,"(%d,%d)%d",&u,&v,&w); addedge(u,v,w); } while(np--){ scanf("%s",str); sscanf(str,"(%d)%d",&v,&w); addedge(src,v,w); } while(nc--){ scanf("%s",str); sscanf(str,"(%d)%d",&u,&w); addedge(u,des,w); } printf("%d\n",Dinic()); } return 0; }
SAP:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int VM=101000; const int EM=500100; const int INF=0x3f3f3f3f; struct Edge{ int to,nxt; int cap; }edge[EM<<1]; int n,np,nc,m,k,cnt,head[VM],map[110][110]; int dep[VM],gap[VM],cur[VM],aug[VM],pre[VM]; void addedge(int cu,int cv,int cw){ edge[cnt].to=cv; edge[cnt].cap=cw; edge[cnt].nxt=head[cu]; head[cu]=cnt++; edge[cnt].to=cu; edge[cnt].cap=0; edge[cnt].nxt=head[cv]; head[cv]=cnt++; } int src,des; int SAP(int n){ int max_flow=0,u=src,v; int id,mindep; aug[src]=INF; pre[src]=-1; memset(dep,0,sizeof(dep)); memset(gap,0,sizeof(gap)); gap[0]=n; for(int i=0;i<=n;i++) cur[i]=head[i]; // 初始化当前弧为第一条弧 while(dep[src]<n){ int flag=0; if(u==des){ max_flow+=aug[des]; for(v=pre[des];v!=-1;v=pre[v]){ // 路径回溯更新残留网络 id=cur[v]; edge[id].cap-=aug[des]; edge[id^1].cap+=aug[des]; aug[v]-=aug[des]; // 修改可增广量，以后会用到 if(edge[id].cap==0) // 不回退到源点，仅回退到容量为0的弧的弧尾 u=v; } } for(int i=cur[u];i!=-1;i=edge[i].nxt){ v=edge[i].to; // 从当前弧开始查找允许弧 if(edge[i].cap>0 && dep[u]==dep[v]+1){ // 找到允许弧 flag=1; pre[v]=u; cur[u]=i; aug[v]=min(aug[u],edge[i].cap); u=v; break; } } if(!flag){ if(--gap[dep[u]]==0) /* gap优化，层次树出现断层则结束算法 */ break; mindep=n; cur[u]=head[u]; for(int i=head[u];i!=-1;i=edge[i].nxt){ v=edge[i].to; if(edge[i].cap>0 && dep[v]<mindep){ mindep=dep[v]; cur[u]=i; // 修改标号的同时修改当前弧 } } dep[u]=mindep+1; gap[dep[u]]++; if(u!=src) // 回溯继续寻找允许弧 u=pre[u]; } } return max_flow; } int main(){ //freopen("input.txt","r",stdin); char str[30]; while(~scanf("%d%d%d%d",&n,&np,&nc,&m)){ cnt=0; memset(head,-1,sizeof(head)); src=n, des=n+1; int u,v,w; while(m--){ scanf("%s",str); sscanf(str,"(%d,%d)%d",&u,&v,&w); addedge(u,v,w); } while(np--){ scanf("%s",str); sscanf(str,"(%d)%d",&v,&w); addedge(src,v,w); } while(nc--){ scanf("%s",str); sscanf(str,"(%d)%d",&u,&w); addedge(u,des,w); } printf("%d\n",SAP(des+1)); } return 0; }
