Power Network
| Time Limit: 2000MS | Memory Limit: 32768K | |
| Total Submissions: 20026 | Accepted: 10539 |
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.
//dinic 网络流 增广路法的网络流算法 理解起来比较简单，速度也还不错
// 对于该题 ： 建立自行建立 S 和 T s和所有发电站相连 所有用户和t相连 cap分别为 pmax cmax
// 其它边都按题目所给的连
// 唉、该类题目的难点在于问题建模、、表示偶才刚刚开始呀、、
//Re了2次、、是因为虽然说m<=n^2但是自行加了边 所以M最好>=20400 我开始只给了了20010....;
#include <iostream> #include <math.h> #include <vector> #include <queue> #include <stack> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int M=20510; const int N=110; const int INF=1000000000; struct Eg { int to; int cap; int flow; int next; }edge[M]; int nu; int node[N]; int level[N]; void add(int from,int to,int cap) { edge[nu].to=to; edge[nu].cap=cap; edge[nu].flow=0; edge[nu].next=node[from]; node[from]=nu++; edge[nu].to=from; edge[nu].cap=0; edge[nu].flow=0; edge[nu].next=node[to]; node[to]=nu++; } int n,np,nc,m; bool BFS(int s,int t) { int u,v,e; memset(level,0,sizeof(level)); level[s]=1; queue<int> Q; Q.push(s); while(!Q.empty()) { u=Q.front(); if(u==t) return true; Q.pop(); for(e=node[u];e!=-1;e=edge[e].next) { v=edge[e].to; if(!level[v]&&edge[e].cap-edge[e].flow>0) { level[v]=level[u]+1; Q.push(v); } } } return false; } int DFS(int u,int maxf,int t) { if(u==t) return maxf; int ret=0,f; int v,e; for(e=node[u];e!=-1;e=edge[e].next) { v=edge[e].to; if(level[v]==level[u]+1&&(edge[e].cap-edge[e].flow)>0) { f=DFS(v,min(maxf-ret,edge[e].cap-edge[e].flow),t); ret+=f; edge[e].flow+=f; edge[e^1].flow-=f; if(ret==maxf) return ret; } } return ret; } int main() { int i; int u,v,z; while(scanf("%d",&n)!=EOF) { scanf("%d %d %d",&np,&nc,&m); memset(node,-1,sizeof(node)); int s=n,t=n+1; n++; nu=0; for(i=0;i<m;i++) { while(getchar()!='('); scanf("%d,%d)%d",&u,&v,&z); add(u,v,z); } for(i=0;i<np;i++) { while(getchar()!='('); scanf("%d)%d",&u,&z); add(s,u,z); } for(i=0;i<nc;i++) { while(getchar()!='('); scanf("%d)%d",&u,&z); add(u,t,z); } int ans=0; while(BFS(s,t)) ans+=DFS(s,INF,t); printf("%d\n",ans); } return 0; }
//由于上面的dinic算法太慢、、我就决定写下 ISAP 。。。好吧、、真的快了好多
#include <iostream>
#include <math.h>
#include <vector>
#include <queue>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int M=20510;
const int N=110;
const int INF=1000000000;
struct Eg
{
int to;
int cap;
int next;
}edge[M];
int nu;
int node[N];
void add(int from,int to,int cap)
{
edge[nu].to=to;
edge[nu].cap=cap;
edge[nu].next=node[from];
node[from]=nu++;
edge[nu].to=from;
edge[nu].cap=0;
edge[nu].next=node[to];
node[to]=nu++;
}
int n,np,nc,m;
int h[N],num[N],cur[N],pre[N];
void rev_bfs(int s,int t) //开始没写这个的、、直接照着书上的写，就是让num[0]=N 、、坑爹的就是一直错、、不知道为什么
{
memset(h,0,sizeof(h));
memset(num,0,sizeof(num));
queue<int> Q;
Q.push(t);
int i,e,u,v;
while(!Q.empty())
{
u=Q.front();
Q.pop();
for(e=node[u];e!=-1;e=edge[e].next)
{
v=edge[e].to;
if(!edge[e].cap&&!h[v])
{
h[v]=h[u]+1;
Q.push(v);
}
}
}
for(i=0;i<=t;i++)
num[h[i]]++;
}
int sap(int s,int t)
{
int cur_flow,cur_ans=0;
int i,u,neck,e;
rev_bfs(s,t);
memset(pre,-1,sizeof(pre));
for(i=0;i<=t;i++)
cur[i]=node[i];
u=s;
while(h[s]<=t)
{
if(u==t)
{
cur_flow=INF;
for(i=s;i!=t;i=edge[cur[i]].to)
{
e=cur[i];
if(cur_flow>edge[e].cap)
{
cur_flow=edge[e].cap;
neck=i;
}
}
for(i=s;i!=t;i=edge[cur[i]].to)
{
e=cur[i];
edge[e].cap-=cur_flow;
edge[e^1].cap+=cur_flow;
}
cur_ans+=cur_flow;
u=neck;
}
for(e=cur[u];e!=-1;e=edge[e].next)
if(edge[e].cap&&h[u]==h[edge[e].to]+1)
break;
if(e!=-1)
{
cur[u]=e;
pre[edge[e].to]=u;
u=edge[e].to;
}
else
{
if(--num[h[u]]==0) break;
cur[u]=node[u];
/* int temp=INF; //这段代码应该换成下面的一段代码，原因是在找到路径时找不下去时
for(e=node[u];e!=-1;e=edge[e].next) //这段代码会在那个点停住、、然后、、就等着Over
if(edge[e].cap) //就是因为这个 temp=INF出了问题
temp=min(temp,h[edge[e].to]); //真是教训深刻、、、、
if(temp!=INF)
{
h[u]=temp+1;
++num[h[u]];
if(u!=s) u=pre[u];
}
*/
//错啦 错啦、、、终于找到哪错了、、
int temp=t+1;
for(e=node[u];e!=-1;e=edge[e].next)
if(edge[e].cap)
temp=min(temp,h[edge[e].to]);
h[u]=temp+1;
++num[h[u]];
if(u!=s) u=pre[u];
}
}
return cur_ans;
}
int main()
{
int i;
int u,v,z;
while(scanf("%d",&n)!=EOF)
{
scanf("%d %d %d",&np,&nc,&m);
memset(node,-1,sizeof(node));
int s=n,t=n+1;
// n++;
nu=0;
for(i=0;i<m;i++)
{
while(getchar()!='(');
scanf("%d,%d)%d",&u,&v,&z);
add(u,v,z);
}
for(i=0;i<np;i++)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
add(s,u,z);
}
for(i=0;i<nc;i++)
{
while(getchar()!='(');
scanf("%d)%d",&u,&z);
add(u,t,z);
}
printf("%d\n",sap(s,t));
}
return 0;
}