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Power Network
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 |
題意：
一個電網包含n個結點，這些結點通過電線連接。
結點一共有三種：
（1）電站：產生電能和傳輸電能
（2）消費者：消耗電能，也可傳輸電能
（3）調度站：不產生電能，不消耗電能，只傳輸電能。
現在要求最大消費電能。電站和消費者有多個。
分析：
新加入一個點作爲源點，向每一個電站引一條弧，權值爲最大生產電量
新加入一個點作爲匯點，每個消費者都向它引一條弧，權值爲最大消耗電量。
其他所有點不管是何種都可以看做是調度站，即進入等於輸出。
構圖完成，用最大流算法求解，本題暫用EK算法，複雜度較高。
代碼：
（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找增廣路徑
{
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]) //改進網絡流
{
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;
}對比如下，第一個爲Dinic，第二個爲EK算法，可見Dinic明顯優於EK。不過代碼量也不是一個等級的。
| 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 // 邊
{
int from, to, cap, flow;
};
vector<Edge> EG; // 邊表
vector<int> G[maxn]; // 鄰接表
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() // 增廣
{
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;
}