http://poj.org/problem?id=1459
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1734
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 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. 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 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 Source |
题意:
一共有n个点,其中np个发电站,nc个用户,剩余的是中转站,有m条电缆(有向),电缆上有容量限制,发电站有发电上限,用户有耗电上限,求电网中最大消耗。
分析:
显然是网络最大流,发电站是源点,用户是汇点,建立超级源点与超级汇点,超级源点与发电站连一条有向边,容量为该发电站的发电上限,用户与超级汇点连一条有向边,容量为该用户的耗电上限。
这题我分别用EK算法和Dinic算法实现,发现在本题中Dinic算法的效率比EK算法高了近20倍!而在POJ 2112中,Dinic算法也比EK算法快了7倍多!Dinic简直就是神器啊,以后都用他了。
比较两种算法,EK算法是一次BFS找一条增广路;Dinic是一次BFS建立分层图,在该分层图上多次DFS找出多条增广路,以此减少BFS的次数,从而获得更高的效率。
EK算法实现:
#include<cstdio> #include<iostream> #include<cstdlib> #include<algorithm> #include<ctime> #include<cctype> #include<cmath> #include<string> #include<cstring> #include<stack> #include<queue> #include<list> #include<vector> #include<map> #include<set> #define sqr(x) ((x)*(x)) #define LL long long #define itn int #define INF 0x3f3f3f3f #define PI 3.1415926535897932384626 #define eps 1e-10 #define maxm 23456 #define maxn 107 using namespace std; int fir[maxn]; int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm]; int e_max; int p[maxn],q[maxn],d[maxn]; void add_edge(int _u,int _v,int _w) { int e; e=e_max++; u[e]=_u;v[e]=_v;cap[e]=_w; nex[e]=fir[u[e]];fir[u[e]]=e; e=e_max++; u[e]=_v;v[e]=_u;cap[e]=0; nex[e]=fir[u[e]];fir[u[e]]=e; } int max_flow(int s,int t) { memset(flow,0,sizeof flow); int total_flow=0; for (;;) { memset(d,0,sizeof d); d[s]=INF; int f=0,r=0; q[0]=s; while (f<=r) { int _u=q[f++]; for (int e=fir[_u];~e;e=nex[e]) { if (!d[v[e]] && cap[e]>flow[e]) { q[++r]=v[e]; p[v[e]]=e; d[v[e]]=min(d[u[e]],cap[e]-flow[e]); } } } if (d[t]==0) break; for (int e=p[t];;e=p[u[e]]) { flow[e]+=d[t]; flow[e^1]-=d[t]; if (u[e]==s) break; } total_flow+=d[t]; } return total_flow; } int main() { #ifndef ONLINE_JUDGE freopen("/home/fcbruce/文档/code/t","r",stdin); #endif // ONLINE_JUDGE int n,np,nc,m,_u,_v,_w; while (~scanf("%d %d %d %d",&n,&np,&nc,&m)) { e_max=0; int s=n,t=n+1; memset(fir,-1,sizeof fir); for (int i=0;i<m;i++) { scanf(" (%d,%d)%d",&_u,&_v,&_w); add_edge(_u,_v,_w); } for (int i=0;i<np;i++) { scanf(" (%d)%d",&_u,&_w); add_edge(s,_u,_w); } for (int i=0;i<nc;i++) { scanf(" (%d)%d",&_u,&_w); add_edge(_u,t,_w); } printf("%d\n",max_flow(s,t)); } return 0; }
dinic算法实现:
#include<cstdio> #include<iostream> #include<cstdlib> #include<algorithm> #include<ctime> #include<cctype> #include<cmath> #include<string> #include<cstring> #include<stack> #include<queue> #include<list> #include<vector> #include<map> #include<set> #define sqr(x) ((x)*(x)) #define LL long long #define itn int #define INF 0x3f3f3f3f #define PI 3.1415926535897932384626 #define eps 1e-10 #define maxm 13000 #define maxn 300 using namespace std; int G[maxn][maxn]; int fir[maxn]; int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm]; int e_max; int lv[maxn],iter[maxn]; int q[maxm]; void add_edge(int _u,int _v,int _w) { int e; e=e_max++; u[e]=_u;v[e]=_v;cap[e]=_w; nex[e]=fir[u[e]];fir[u[e]]=e; e=e_max++; u[e]=_v;v[e]=_u;cap[e]=0; nex[e]=fir[u[e]];fir[u[e]]=e; } void dinic_bfs(int s) { int f,r; memset(lv,-1,sizeof lv); q[f=r=0]=s; lv[s]=0; while(f<=r) { int x=q[f++]; for (int e=fir[x];~e;e=nex[e]) { if (cap[e]>flow[e] && lv[v[e]]<0) { lv[v[e]]=lv[u[e]]+1; q[++r]=v[e]; } } } } int dinic_dfs(int _u,int t,int _f) { if (_u==t) return _f; for (int &e=iter[_u];~e;e=nex[e]) { if (cap[e]>flow[e] && lv[_u]<lv[v[e]]) { int _d=dinic_dfs(v[e],t,min(_f,cap[e]-flow[e])); if (_d>0) { flow[e]+=_d; flow[e^1]-=_d; return _d; } } } return 0; } int max_flow(int s,int t) { memset(flow,0,sizeof flow); int total_flow=0; for (;;) { dinic_bfs(s); if (lv[t]<0) return total_flow; memcpy(iter,fir,sizeof iter); int _f; while ((_f=dinic_dfs(s,t,INF))>0) total_flow+=_f; } return total_flow; } int main() { #ifndef ONLINE_JUDGE freopen("/home/fcbruce/文档/code/t","r",stdin); #endif // ONLINE_JUDGE int l=INF,r=0,K,C,M; scanf("%d %d %d",&K,&C,&M); for (int i=1;i<=K+C;i++) for (int j=1;j<=K+C;j++) scanf("%d",&G[i][j]),G[i][j]=G[i][j]?G[i][j]:INF; for (int k=1;k<=K+C;k++) for (int i=1;i<=K+C;i++) for (int j=1;j<=K+C;j++) G[i][j]=min(G[i][k]+G[k][j],G[i][j]); int ans=-1; // printf("%d %d\n",l,r); l=0;r=INF; while (l<=r) { int mid=l+r>>1; e_max=0; memset(fir,-1,sizeof fir); for (int i=K+1;i<=K+C;i++) add_edge(0,i,1); for (int i=1;i<=K;i++) add_edge(i,K+C+1,M); for (int i=K+1;i<=K+C;i++) for (int j=1;j<=K;j++) if (G[i][j]<=mid) add_edge(i,j,1); // printf("mid=%d\n",mid); if (max_flow(0,K+C+1)==C) { ans=mid; r=mid-1; } else { l=mid+1; } } printf("%d\n",ans); return 0; }
POJ 1459 & ZOJ 1734 Power Network (网络最大流)