POJ1459 Power Network（网络最大流）
Power Network
| Time Limit: 2000MS | Memory Limit: 32768K | |
| Total Submissions: 27229 | Accepted: 14151 |
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.
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.
/* EdmondsKarp Memory 320K Time 1282MS 和没过没什么区别 */ #include <iostream> #include <queue> using namespace std; #define min(a,b) (a<b?a:b) #define MAXV 105 #define MAXINT INT_MAX typedef struct{ int flow; //流量 int capacity; //最大容量值 }maps; maps map[MAXV][MAXV]; int vertime; //顶点总数 int nedges; //边的总数 int power_stations; //发电站总数 int consumers; //消费者总数 int maxflow; //最大流 int sp,fp; //标记源点与汇点 int parent[MAXV]; //用于bfs寻找路径 int bfs(int start,int end){ int a[MAXV],i,v; queue <int>q; memset(a,0,sizeof(a)); memset(parent,-1,sizeof(parent)); q.push(start); a[start]=MAXINT; while(!q.empty()){ v=q.front();q.pop(); for(i=1;i<=vertime;i++){ if(!a[i] && map[v][i].capacity>map[v][i].flow){ q.push(i); parent[i]=v; a[i]=min(a[v],map[v][i].capacity-map[v][i].flow); } } if(v==end) break; } return a[end]; } void EdmondsKarp(){ int i,tmp; maxflow=0; while(tmp=bfs(sp,fp)){ for(i=fp;i!=sp;i=parent[i]){ map[i][parent[i]].flow-=tmp; //更新反向流 map[parent[i]][i].flow+=tmp; //更新正向流 } maxflow+=tmp; } } int main(){ int i; int x,y,z; char ch; while(scanf("%d%d%d%d", &vertime, &power_stations,&consumers,&nedges)!= EOF){ //Init memset(map,0,sizeof(map)); //Read Gragh for(i=1;i<=nedges;i++){ //设置读图从1开始 cin>>ch>>x>>ch>>y>>ch>>z; map[x+1][y+1].capacity=z; } //Build Gragh //建立超级源点指向所有的发电站 sp=vertime+1;fp=vertime+2;vertime+=2; for (i=1; i<=power_stations; i++){ cin>>ch>>x>>ch>>y; map[sp][x+1].capacity=y; } //建立超级汇点，使所有消费者指向它 for (i=1; i<=consumers; i++){ cin>>ch>>x>>ch>>y; map[x+1][fp].capacity=y; } EdmondsKarp(); printf("%d\n",maxflow); } return 0; }
/* dinic Memory 320K Time 563MS */ #include <iostream> #include <queue> using namespace std; #define min(a,b) (a<b?a:b) #define MAXV 105 #define MAXINT INT_MAX typedef struct{ int flow; //流量 int capacity; //最大容量值 }maps; maps map[MAXV][MAXV]; int dis[MAXV]; //用于dinic分层 int vertime; //顶点总数 int nedges; //边的总数 int power_stations; //发电站总数 int consumers; //消费者总数 int maxflow; //最大流 int sp,fp; //标记源点与汇点 bool bfs(){ int v,i; queue <int>q; memset(dis,0,sizeof(dis)); q.push(sp); dis[sp]=1; while(!q.empty()){ v=q.front();q.pop(); for(i=1;i<=vertime;i++) if(!dis[i] && map[v][i].capacity>map[v][i].flow){ q.push(i); dis[i]=dis[v]+1; } if(v==fp) return 1; } return 0; } int dfs(int cur,int cp){ if(cur==fp) return cp; int tmp=cp,t; for(int i=1;i<=vertime;i++) if(dis[i]==dis[cur]+1 && tmp && map[cur][i].capacity>map[cur][i].flow){ t=dfs(i,min(map[cur][i].capacity-map[cur][i].flow,tmp)); map[cur][i].flow+=t; map[i][cur].flow-=t; tmp-=t; } return cp-tmp; } void dinic(){ maxflow=0; while(bfs()) maxflow+=dfs(sp,MAXINT); } int main(){ int i; int x,y,z; char ch; while(scanf("%d%d%d%d", &vertime, &power_stations,&consumers,&nedges)!= EOF){ //Init memset(map,0,sizeof(map)); //Read Gragh for(i=1;i<=nedges;i++){ //设置读图从1开始 cin>>ch>>x>>ch>>y>>ch>>z; map[x+1][y+1].capacity=z; } //Build Gragh //建立超级源点指向所有的发电站 sp=vertime+1;fp=vertime+2;vertime+=2; for (i=1; i<=power_stations; i++){ cin>>ch>>x>>ch>>y; map[sp][x+1].capacity=y; } //建立超级汇点，使所有消费者指向它 for (i=1; i<=consumers; i++){ cin>>ch>>x>>ch>>y; map[x+1][fp].capacity=y; } dinic(); printf("%d\n",maxflow); } return 0; }
/* sap Memory 328K Time 454MS */ #include <iostream> #include <queue> using namespace std; #define MAXV 110 #define INF 1<<29 #define min(a,b) (a>b?b:a) int n,c[MAXV][MAXV],r[MAXV][MAXV],source,sink; int dis[MAXV],maxflow; void bfs(){ int v,i; queue <int>q; memset(dis,0,sizeof(dis)); q.push(sink); while(!q.empty()){ v=q.front();q.pop(); for(i=0;i<=sink;i++){ if(!dis[i] && c[i][v]>0){ dis[i] = dis[v] +1; q.push(i); } } } } void sap(){ int top=source,pre[MAXV],i,j,low[MAXV]; bfs(); //分层 memset(low,0,sizeof(low)); //保存路径的最小流量 while(dis[source]<n){ low[source]=INF; for(i=0;i<=sink;i++){ //找到一条允许弧 if(r[top][i]>0 && dis[top]==dis[i] +1) break; } if(i<=sink){ //找到了 low[i]=min(r[top][i],low[top]); //更新最小流量 pre[i]=top;top=i; //记录增广路径 if(top==sink){ //找到一条增广路径更新残量 maxflow += low[sink]; j = top; while(j != source){ i=pre[j]; r[i][j]-=low[sink]; r[j][i]+=low[sink]; j=i; } top=source; //再从头找一条增广路径 memset(low,0,sizeof(low)); } } else{ //找不到这样一条允许弧更新距离数组 int mindis=INF; for(j=0;j <=sink;j++){ if(r[top][j]>0 && mindis>dis[j] +1) mindis=dis[j] +1; } dis[top]=mindis; if(top!=source) top=pre[top]; } } } int main(){ int i,nedges,power_stations,consumers; int x,y,z; char ch; while(scanf("%d%d%d%d", &n, &power_stations,&consumers,&nedges)!= EOF){ //Init memset(r,0,sizeof(r)); memset(c,0,sizeof(c)); source=0;sink=n+1;n+=2;maxflow=0; //Read Gragh for(i=1;i<=nedges;i++){ //设置读图从1开始 cin>>ch>>x>>ch>>y>>ch>>z; c[x+1][y+1]=r[x+1][y+1]=z; } //Build Gragh //建立超级源点指向所有的发电站 for (i=1;i<=power_stations;i++){ cin>>ch>>x>>ch>>y; c[source][x+1]=r[source][x+1]=y; } //建立超级汇点，使所有消费者指向它 for (i=1;i<=consumers;i++){ cin>>ch>>x>>ch>>y; c[x+1][sink]=r[x+1][sink]=y; } sap(); printf("%d\n",maxflow); } return 0; }
sap+分层+gap优化 Memory 328K Time 438MS */ #include <iostream> #include <queue> using namespace std; #define MAXV 110 #define INF 1<<29 #define min(a,b) (a>b?b:a) int n,c[MAXV][MAXV],r[MAXV][MAXV],source,sink; int dis[MAXV],maxflow,gap[MAXV]; void bfs(){ int v,i; queue <int>q; memset(dis,0,sizeof(dis)); memset(gap,0,sizeof(gap)); gap[0]++; q.push(sink); while(!q.empty()){ v=q.front();q.pop(); for(i=0;i<=sink;i++){ if(!dis[i] && c[i][v]>0){ dis[i] = dis[v] +1; gap[dis[i]]++; q.push(i); } } } } void sap(){ int top=source,pre[MAXV],i,j,low[MAXV]; bfs(); //分层 memset(low,0,sizeof(low)); //保存路径的最小流量 while(dis[source]<n){ low[source]=INF; for(i=0;i<=sink;i++){ //找到一条允许弧 if(r[top][i]>0 && dis[top]==dis[i]+1 && dis[i]>=0) break; } if(i<=sink){ //找到了 low[i]=min(r[top][i],low[top]); //更新最小流量 pre[i]=top;top=i; //记录增广路径 if(top==sink){ //找到一条增广路径更新残量 maxflow += low[sink]; j = top; while(j != source){ i=pre[j]; r[i][j]-=low[sink]; r[j][i]+=low[sink]; j=i; } top=source; //再从头找一条增广路径 memset(low,0,sizeof(low)); } } else{ //找不到这样一条允许弧更新距离数组 int mindis=INF; for(j=0;j <=sink;j++){ if(r[top][j]>0 && mindis>dis[j] +1 && dis[j]>=0) mindis=dis[j] +1; } gap[dis[top]]--; if (gap[dis[top]] ==0) break; gap[mindis]++; dis[top]=mindis; if(top!=source) top=pre[top]; } } } int main(){ int i,nedges,power_stations,consumers; int x,y,z; char ch; while(scanf("%d%d%d%d", &n, &power_stations,&consumers,&nedges)!= EOF){ //Init memset(r,0,sizeof(r)); memset(c,0,sizeof(c)); source=0;sink=n+1;n+=2;maxflow=0; //Read Gragh for(i=1;i<=nedges;i++){ //设置读图从1开始 cin>>ch>>x>>ch>>y>>ch>>z; c[x+1][y+1]=r[x+1][y+1]=z; } //Build Gragh //建立超级源点指向所有的发电站 for (i=1;i<=power_stations;i++){ cin>>ch>>x>>ch>>y; c[source][x+1]=r[source][x+1]=y; } //建立超级汇点，使所有消费者指向它 for (i=1;i<=consumers;i++){ cin>>ch>>x>>ch>>y; c[x+1][sink]=r[x+1][sink]=y; } sap(); printf("%d\n",maxflow); } return 0; }
