ZOJ 1698 (最大流入門）
Power NetworkTime Limit:5000MS Memory Limit:32768KB 64bit IO Format:%lld
& %llu
SubmitStatusPracticeZOJ
1734
Appoint description:
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=sum of c(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.
There are several data sets in the input text file. 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.
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.
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.
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
EK臨接矩陣版：
#include
#include
#include
#include
#include
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=150;//點數的最大值
const int MAXM=20500;//邊數的最大值
int n;
int flow[MAXN][MAXN],cap[MAXN][MAXN],p[MAXN],a[MAXN];
void Init()
{
memset(flow,0,sizeof flow);
memset(cap,0,sizeof cap);
}
int Ek(int s,int t){
queueq;
int f=0;
while(1){
memset(a,0,sizeof a);
while(!q.empty())q.pop();
a[s]=INF;
q.push(s);
while(!q.empty()){
int u=q.front();q.pop();
for(int v=0;v<=n+1;v++)if(!a[v]&&cap[u][v]>flow[u][v]){
q.push(v);
p[v]=u;
a[v]=min(a[u],cap[u][v]-flow[u][v]);
}
}
if(a[t]==0)return f;
int x=t;
while(x!=s){
flow[p[x]][x]+=a[t];
flow[x][p[x]]-=a[t];
x=p[x];
}
f+=a[t];
}
return f;
}
void addedge(int u,int v,int w){
cap[u][v]+=w;
}
int main()//多源多匯點，在前面加個源點，後面加個匯點，轉成單源單匯點
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int start,end;
int np,nc,m;
int u,v,z;
while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
{
Init();
while(m--)
{
while(getchar()!='(');
scanf("%d,%d)%d",&u,&v,&z);
u++;v++;
addedge(u,v,z);
}
while(np--)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
u++;
addedge(0,u,z);
}
while(nc--)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
u++;
addedge(u,n+1,z);
}
start=0;
end=n+1;
int ans=Ek(start,end);
cout<
EK臨接表版：
#include
#include
#include
#include
#include
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=150;//點數的最大值
const int MAXM=20500;//邊數的最大值
int n,a[MAXN],p[MAXN];
struct Edge
{
int from,to,cap,flow;
};
std::vectoredges;
std::vectorG[MAXN];
void addedge(int u,int v,int w){
edges.push_back((Edge){u,v,w,0});
edges.push_back((Edge){v,u,0,0});
int m=edges.size();
G[u].push_back(m-2);
G[v].push_back(m-1);
}
int Ek(int s,int t){
queueq;
int f=0;
while(1){
memset(a,0,sizeof a);
a[s]=INF;
q.push(s);
while(!q.empty()){
int u=q.front();q.pop();
for(int i=0;ie.flow){
q.push(e.to);
p[e.to]=G[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
}
}
}
if(a[t]==0)return f;
int x=t;
while(x!=s){
edges[p[x]].flow+=a[t];
edges[p[x]^1].flow-=a[t];
x=edges[p[x]].from;
}
f+=a[t];
}
return f;
}
int main()//多源多匯點，在前面加個源點，後面加個匯點，轉成單源單匯點
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int start,end;
int np,nc,m;
int u,v,z;
while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
{
edges.clear();
for(int i=0;i<=n+1;i++)G[i].clear();
while(m--)
{
while(getchar()!='(');
scanf("%d,%d)%d",&u,&v,&z);
u++;v++;
addedge(u,v,z);
}
while(np--)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
u++;
addedge(0,u,z);
}
while(nc--)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
u++;
addedge(u,n+1,z);
}
start=0;
end=n+1;
int ans=Ek(start,end);
cout<
dinic算法版：
#include
#include
#include
#include
#include
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=150;//點數的最大值
const int MAXM=20500;//邊數的最大值
int n,a[MAXN],p[MAXN];
int cur[MAXN],d[MAXN],vis[MAXN];
struct Edge
{
int from,to,cap,flow;
};
std::vectoredges;
std::vectorG[MAXN];
void addedge(int u,int v,int w){
edges.push_back((Edge){u,v,w,0});
edges.push_back((Edge){v,u,0,0});
int m=edges.size();
G[u].push_back(m-2);
G[v].push_back(m-1);
}
int bfs(int s,int t)
{
memset(vis,0,sizeof vis);
queueq;
q.push(s);
vis[s]=1;
d[s]=0;
while(!q.empty()){
int u=q.front();q.pop();
for(int i=0;ie.flow){
q.push(e.to);
vis[e.to]=1;
d[e.to]=d[u]+1;
}
}
}
return vis[t];
}
int dfs(int x,int a,int t){
if(x==t||a==0)return a;
int flow=0,f=0;
for(int &i=cur[x];i0){
e.flow+=f;
edges[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if(a==0)break;
}
}
return flow;
}
int dinic(int s,int t){
int flow=0;
while(bfs(s,t)){
memset(cur,0,sizeof cur);
flow+=dfs(s,INF,t);
}
return flow;
}
int main()//多源多匯點，在前面加個源點，後面加個匯點，轉成單源單匯點
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int start,end;
int np,nc,m;
int u,v,z;
while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
{
edges.clear();
for(int i=0;i<=n+1;i++)G[i].clear();
while(m--)
{
while(getchar()!='(');
scanf("%d,%d)%d",&u,&v,&z);
u++;v++;
addedge(u,v,z);
}
while(np--)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
u++;
addedge(0,u,z);
}
while(nc--)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
u++;
addedge(u,n+1,z);
}
start=0;
end=n+1;
int ans=dinic(start,end);
cout<
