POJ 1459 Power Network 经典网络流构图问题 最大流，EK算法、Dinic算法、ISAP算法

题目链接：POJ 1459 Power Network

Power Network Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 23347 Accepted: 12231 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)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. Source Southeastern Europe 2003

题意：
一个电网包含n个结点，这些结点通过电线连接。

结点一共有三种：

（1）电站：产生电能和传输电能

（2）消费者：消耗电能，也可传输电能

（3）调度站：不产生电能，不消耗电能，只传输电能。

现在要求最大消费电能。电站和消费者有多个。

分析：

新加入一个点作为源点，向每一个电站引一条弧，权值为最大生产电量

新加入一个点作为汇点，每个消费者都向它引一条弧，权值为最大消耗电量。

其他所有点不管是何种都可以看做是调度站，即进入等于输出。

构图完成，用最大流算法求解，本题暂用EK算法，复杂度较高。

代码：

（1）EK算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define maxn 110
#define INF 0x3f3f3f3f
int ans, s, t, n;
int a[maxn], pre[maxn];
int flow[maxn][maxn];
int cap[maxn][maxn];

void Edmonds_Karp()
{
queue<int> q;
memset(flow, 0, sizeof(flow));
ans = 0;
while(1)
{
memset(a, 0, sizeof(a));
a[s] = INF;
q.push(s);
while(!q.empty())   //bfs找增广路径
{
int u = q.front();
q.pop();
for(int v = 0; v < n; v++)
if(!a[v] && cap[u][v] > flow[u][v])
{
pre[v] = u;
q.push(v);
a[v] = min(a[u], cap[u][v]-flow[u][v]);
}
}
if(a[t] == 0) break;
for(int u = t; u != s; u = pre[u])  //改进网络流
{
flow[pre[u]][u] += a[t];
flow[u][pre[u]] -= a[t];
}
ans += a[t];
}
}

int main()
{
//freopen("poj_1459.txt", "r", stdin);
int m, u, v, c, np, nc;
while(~scanf("%d%d%d%d", &n, &np, &nc, &m))
{
s = n, t = n+1;
n += 2;
//cout << s << " " << t << " " << n << endl;
memset(cap, 0, sizeof(cap));
while(m--)
{
while(getchar() !='(');
scanf("%d,%d)%d", &u, &v, &c);
cap[u][v] = c;
}
while(np--)
{
while(getchar() != '(');
scanf("%d)%d", &v, &c);
cap[s][v] = c;
}
while(nc--)
{
while(getchar() != '(');
scanf("%d)%d", &u, &c);
cap[u][t] = c;
}
/*
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
cout << cap[i][j] << " ";
cout << endl;
}
*/
Edmonds_Karp();
printf("%d\n", ans);
}
return 0;
}

（2）Dinic算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define maxn 110
#define INF 0x3f3f3f3f

struct Edge
{
int from, to, cap;
};

vector<Edge> EG;
vector<int> G[maxn];
int n, s, t, ans, d[maxn], cur[maxn];

void addEdge(int from, int to, int cap)
{
EG.push_back((Edge){from, to, cap});
EG.push_back((Edge){to, from, 0});
int x = EG.size();
G[from].push_back(x-2);
G[to].push_back(x-1);
}

bool bfs()
{
memset(d, -1, sizeof(d));
queue<int> q;
q.push(s);
d[s] = 0;
while(!q.empty())
{
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(d[e.to] == -1 && e.cap > 0)
{
d[e.to] = d[x]+1;
q.push(e.to);
}
}
}
return (d[t]!=-1);
}

int dfs(int x, int a)
{
if(x == t || a == 0) return a;
int flow = 0, f;
for(int& i = cur[x]; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(d[x]+1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0)
{
e.cap -= f;
EG[G[x][i]^1].cap += f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
void Dinic()
{
ans = 0;
while(bfs())
{
memset(cur, 0, sizeof(cur));
ans += dfs(s, INF);
}
}
int main()
{
//freopen("poj_1459.txt", "r", stdin);
int m, u, v, c, np, nc;
while(~scanf("%d%d%d%d", &n, &np, &nc, &m))
{
s = n, t = n+1;
n += 2;
while(m--)
{
while(getchar() !='(');
scanf("%d,%d)%d", &u, &v, &c);
addEdge(u, v, c);
}
while(np--)
{
while(getchar() != '(');
scanf("%d)%d", &v, &c);
addEdge(s, v, c);
}
while(nc--)
{
while(getchar() != '(');
scanf("%d)%d", &u, &c);
addEdge(u, t, c);
}
Dinic();
printf("%d\n", ans);
EG.clear();
for(int i = 0; i <= n; ++i)
G[i].clear();
}
return 0;
}

对比如下，第一个为Dinic，第二个为EK算法，可见Dinic明显优于EK。不过代码量也不是一个等级的。

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
13472137	dzk_acmer	1459	Accepted	1360K	63MS	G++	2293B	2014-09-24 01:56:27
13469601	dzk_acmer	1459	Accepted	760K	625MS	G++	2005B	2014-09-23 13:56:20

（3）ISAP算法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;

#define maxn 110
#define INF 0x3f3f3f3f

struct Edge // 边
{
int from, to, cap, flow;
};

vector<Edge> EG;    // 边表
vector<int> G[maxn];    // 邻接表
int n, s, t, ans, d[maxn], cur[maxn], p[maxn], num[maxn];
bool vis[maxn];

void addEdge(int from, int to, int cap)
{
EG.push_back((Edge){from, to, cap, 0});
EG.push_back((Edge){to, from, 0, 0});
int x = EG.size();
G[from].push_back(x-2);
G[to].push_back(x-1);
}

void bfs()  // 逆向bfs
{
memset(vis, false, sizeof(vis));
queue<int> q;
vis[t] = true;
d[t] = 0;
q.push(t);
while(!q.empty())
{
int x = q.front();
q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]^1];
if(!vis[e.from] && e.cap > e.flow)
{
vis[e.from] = true;
d[e.from] = d[x]+1;
q.push(e.from);
}
}
}
}

int augment()   // 增广
{
int x = t, a = INF;
while(x != s)
{
Edge& e = EG[p[x]];
a = min(a, e.cap-e.flow);
x = EG[p[x]].from;
}
x = t;
while(x != s)
{
EG[p[x]].flow += a;
EG[p[x]^1].flow -= a;
x = EG[p[x]].from;
}
return a;
}
void ISAP()     // ISAP
{
ans =0;
bfs();
memset(num, 0, sizeof(num));
for(int i = 0; i < n; i++)
num[d[i]]++;
int x = s;
memset(cur, 0, sizeof(cur));
while(d[s] < n)
{
if(x == t)
{
ans += augment();
x = s;
}
bool flag = false;
for(int i = cur[x]; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(e.cap > e.flow && d[x] == d[e.to]+1) // advance
{
flag = true;
p[e.to] = G[x][i];
cur[x] = i;
x = e.to;
break;
}
}
if(!flag)   // retreat
{
int m = n-1;
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = EG[G[x][i]];
if(e.cap > e.flow)
m = min(m, d[e.to]);
}
if(--num[d[x]] == 0) break;
num[d[x] = m+1]++;
cur[x] = 0;
if(x != s)
x = EG[p[x]].from;
}
}
}
int main()
{
//freopen("poj_1459.txt", "r", stdin);
int m, u, v, c, np, nc;
while(~scanf("%d%d%d%d", &n, &np, &nc, &m))
{
s = n, t = n+1;
n += 2;
while(m--)
{
while(getchar() !='(');
scanf("%d,%d)%d", &u, &v, &c);
addEdge(u, v, c);
}
while(np--)
{
while(getchar() != '(');
scanf("%d)%d", &v, &c);
addEdge(s, v, c);
}
while(nc--)
{
while(getchar() != '(');
scanf("%d)%d", &u, &c);
addEdge(u, t, c);
}
//Dinic();
ISAP();
printf("%d\n", ans);
EG.clear();
for(int i = 0; i < n; ++i)
G[i].clear();
}
return 0;
}