題目鏈接：https://vjudge.net/problem/POJ-1459

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Power Network

Time Limit:?2000MS	?	Memory Limit:?32768K
Total Submissions:?29270	?	Accepted:?15191

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) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(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) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(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 p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(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 l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(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 c_max(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

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題意：

有np個供電站（只供電不用電）、nc個用電站（只用電不供電），以及n-np-nc個中轉站（既不供電也不用電），且已經知道這些站的連接關系，問單位時間最多能消耗多少的電？

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題解：

最大流問題。

1.建立超級源點，超級源點與每個供電站相連，且邊的容量為供電站的最大供電量，表明流經此供電站的電量最多只能為自身的供電量。

2.建立超級匯點，每個用電站與超級匯點相連，且邊的容量為用電站的最大用電量，表明流經此用電站的電量最多只能為自身的消耗量。

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代碼如下：

1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int mod = 1e9+7; 17 const int MAXN = 1e2+10; 18 19 int maze[MAXN][MAXN]; 20 int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN]; 21 int flow[MAXN][MAXN]; 22 23 int sap(int start, int end, int nodenum) 24 { 25 memset(cur, 0, sizeof(cur)); 26 memset(dis, 0, sizeof(dis)); 27 memset(gap, 0, sizeof(gap)); 28 memset(flow, 0, sizeof(flow)); 29 int u = pre[start] = start, maxflow = 0, aug = INF; 30 gap[0] = nodenum; 31 32 while(dis[start]<nodenum) 33 { 34 loop: 35 for(int v = cur[u]; v<nodenum; v++) 36 if(maze[u][v]-flow[u][v]>0 && dis[u] == dis[v]+1) 37 { 38 aug = min(aug, maze[u][v]-flow[u][v]); 39 pre[v] = u; 40 u = cur[u] = v; 41 if(v==end) 42 { 43 maxflow += aug; 44 for(u = pre[u]; v!=start; v = u, u = pre[u]) 45 { 46 flow[u][v] += aug; 47 flow[v][u] -= aug; 48 } 49 aug = INF; 50 } 51 goto loop; 52 } 53 54 int mindis = nodenum-1; 55 for(int v = 0; v<nodenum; v++) 56 if(maze[u][v]-flow[u][v]>0 && mindis>dis[v]) 57 { 58 cur[u] = v; 59 mindis = dis[v]; 60 } 61 if((--gap[dis[u]])==0) break; 62 gap[dis[u]=mindis+1]++; 63 u = pre[u]; 64 } 65 return maxflow; 66 } 67 68 int main() 69 { 70 int n, np, nc, m; 71 while(scanf("%d%d%d%d", &n,&np,&nc,&m)!=EOF) 72 { 73 int start = n, end = n+1; 74 memset(maze, 0, sizeof(maze)); 75 for(int i = 1; i<=m; i++) 76 { 77 int u, v, w; 78 while(getchar()!='('); 79 scanf("%d,%d)%d",&u,&v,&w); 80 maze[u][v] = w; 81 } 82 83 for(int i = 1; i<=np; i++) 84 { 85 int id, p; 86 while(getchar()!='('); 87 scanf("%d)%d", &id,&p); 88 maze[start][id] = p; 89 } 90 91 for(int i = 1; i<=nc; i++) 92 { 93 int id, p; 94 while(getchar()!='('); 95 scanf("%d)%d", &id,&p); 96 maze[id][end] = p; 97 } 98 99 cout<< sap(start, end, n+2) <<endl; 100 } 101 } View Code

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轉載于:https://www.cnblogs.com/DOLFAMINGO/p/8081368.html

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