POJ 1459 & ZOJ 1734 Power Network (網路最大流)

標籤：algorithm   圖論   網路流   

http://poj.org/problem?id=1459

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1734

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 22674   Accepted: 11880 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)207 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 156 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. Source Southeastern Europe 2003

題意：

一共有n個點，其中np個發電站，nc個使用者，剩餘的是中轉站，有m條電纜（有向），電纜上有容量限制，發電站有發電上限，使用者有耗電上限，求電網中最大消耗。

分析：

顯然是網路最大流，發電站是源點，使用者是匯點，建立超級源點與超級匯點，超級源點與發電站連一條有向邊，容量為該發電站的發電上限，使用者與超級匯點連一條有向邊，容量為該使用者的耗電上限。

這題我分別用EK演算法和Dinic演算法實現，發現在本題中Dinic演算法的效率比EK演算法高了近20倍！而在POJ 2112中，Dinic演算法也比EK演算法快了7倍多！Dinic簡直就是神器啊，以後都用他了。

比較兩種演算法，EK演算法是一次BFS找一條增廣路;Dinic是一次BFS建立分層圖，在該分層圖上多次DFS找出多條增廣路，以此減少BFS的次數，從而獲得更高的效率。

EK演算法實現：

#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 107using 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/文檔/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演算法實現：

#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 300using 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/文檔/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;}