POJ 刷题系列：1459. Power Network

传送门：1459. Power Network

题意：

给你n个点，其中有np个是能提供电力的点，nc个是能消费电力的点，剩下的点(n-np-nc)是中转战即不提供电力也不消费电力，点与点之间是有线路存在的，有m条线路，每条线路有最多运载限定。undefined 前4个数据就是有n个点，np个供电点,nc个消费点,m条线路，接来题目先给出的是m条线路的数据，(起点,终点)最多运载量,然后是np个供电点的数据(供电点)最多供电量，接着就是nc个消费点的数据(消费点)最多消费电量。题目要我们求出给定的图最大能消费的总电量（就是求最大流）.

思路：

最大流模版题，不多说。虚拟出源点0和汇点n 1,将发电厂与源点相连，将用户和汇点相连，然后便可以直接求最大流。采用DINIC算法。

代码语言：javascript复制

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201801/P1459.txt";
    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

    static int nextInt() throws IOException {
        in.nextToken();
        return (int)in.nval;
    }

    static int next() throws IOException {
        in.nextToken();
        in.nextToken();
        return (int) in.nval;
    }


    public static void main(String[] args) throws IOException {
        new Main().read();
    }

    class Edge{
        int to;
        int cap;
        int rev;

        Edge(int to, int cap, int rev){
            this.to  = to;
            this.cap = cap;
            this.rev = rev;
        }

        @Override
        public String toString() {
            return "("   to   ", "   cap   ","   rev   ")";
        }
    }

    static final int INF = 0x3f3f3f3f;

    List<Edge>[] graph;
    int[] level;

    void addEdge(int from, int to, int cap) {
        graph[from].add(new Edge(to, cap, graph[to].size()));
        graph[to].add(new Edge(from, 0, graph[from].size() - 1));
    }

    void bfs(int s, int to) {
        Arrays.fill(level, -1);
        Queue<Integer> q = new ArrayDeque<Integer>();
        level[s] = 0;
        q.offer(s);
        while (!q.isEmpty()) {
            int now = q.poll();
            for (Edge e : graph[now]) {
                if (level[e.to] < 0 && e.cap > 0) {
                    level[e.to] = level[now]   1;
                    q.offer(e.to);
                }
            }
        }
    }

    int dfs(int s, int v, int f, boolean[] vis) {
        if (s == v) return f;
        vis[s] = true;
        for (Edge e : graph[s]) {
            int to  = e.to;
            if (!vis[to] && e.cap > 0 && level[s]   1 == level[to]) {
                int d = dfs(to, v, Math.min(f, e.cap), vis);
                if (d > 0) {
                    e.cap -= d;
                    graph[to].get(e.rev).cap  = d;
                    return d;
                }
            }
        }
        return 0;
    }

    int dinic(int s, int to, int V) {
        int flow = 0;
        for(;;) {
            bfs(s, to);
            if (level[to] < 0) break;
            int f = 0;
            while ((f = dfs(s, to, INF, new boolean[V])) > 0) flow  = f;
        }
        return flow;
    }

    void read() throws IOException {
        while(in.nextToken() != StreamTokenizer.TT_EOF) {
            int n  = (int) in.nval;  // n 个顶点
            int np = nextInt();  // power stations
            int nc = nextInt();  // consumers
            int m  = nextInt();  // power stations line

            graph = new ArrayList[n   2];
            level = new int[n   2];
            for (int i = 0; i < n   2;   i) graph[i] = new ArrayList<Edge>();
            for (int i = 0; i < m;   i) {
                addEdge(next()   1, next()   1, next());
            }

            for (int i = 0; i < np;   i) {
                addEdge(0, next()   1, next());
            }

            for (int i = 0; i < nc;   i) {
                addEdge(next()   1, n   1, next());
            }

            System.out.println(dinic(0, n   1, n   2));
        }
    }
}

network

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