POJ-1459(最大流+EK算法)

Power Network

POJ-1459

• 这题值得思索的就是特殊的输入，如何输入一连串字符。这里采用的方法是根据输入已知的输入格式，事先预定好要接受的数据类型。
• 这里套用的板子也是最大流的模板，但是虽然可以ac但是时间有点卡，所以如果可以的话还是使用高级的算法。

#include<iostream>
#include<algorithm>
#include<fstream>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstring>
using namespace std;
//fstream fin("1.txt");
//streambuf *buf = cin.rdbuf(fin.rdbuf());//用于重定项输入改成，把cin当成fin
const int INF = 1 << 29;
const int maxn = 310;
int n, np, nc, m;
struct Edge {
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
struct EK{
    int m;
    vector<Edge> edges;//这里的边是实际边数的两倍，包括反向边
    vector<int> G[maxn];//邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
    int a[maxn];//a[i]表示起点到i的可改进量
    int p[maxn];//edges数组中的编号，最短路图上p的入弧编号
    void init(int n) {
        for (int i = 0; i <= n; i++)
            G[i].clear();
        edges.clear();
    }
    void AddEdge(int from, int to, int cap) {
        edges.push_back(Edge(from, to, cap, 0));
        edges.push_back(Edge(to, from, 0, 0));//反向弧，容量为0
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }
    int Maxflow(int s, int t) {
        int flow = 0;
        for (;;) {
            memset(a, 0, sizeof(a));
            queue<int> Q;
            Q.push(s);
            a[s] = INF;
            while (!Q.empty()) {
                int x = Q.front();
                Q.pop();
                for (int i = 0; i < G[x].size(); i++) {
                    Edge& e = edges[G[x][i]];
                    if (!a[e.to] && e.cap > e.flow) {
                        p[e.to] = G[x][i];
                        a[e.to] = min(a[x], e.cap - e.flow);
                        Q.push(e.to);
                    }
                }
                if (a[t]) break;
            }
            if (!a[t]) break;
            for (int u = t; u != s; u = edges[p[u]].from) {
                edges[p[u]].flow += a[t];
                edges[p[u] ^ 1].flow -= a[t];
            }
            flow += a[t];
        }
        return flow;
    }
}ek;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    char rubbish;
    int first, next;
    int value;
    while (cin >> n >> np >> nc >> m)
    {
        ek.init(n+1);
        int s = n;
        int t = n + 1;
        for (int i = 0; i < m; i++)
        {
            cin >> rubbish;
            cin >> first;
            cin >> rubbish;
            cin >> next;
            cin >> rubbish;
            cin >> value;
            //cout<<rubbish<<" "<<first<<" "<<rubbish<<" "<<next<<" "<<rubbish<<" "<<value<<endl;
            if (first == next)
                continue;
            //getchar();
            ek.AddEdge(first,next,value);
        }
        for (int i = 0; i < np; i++)
        {
            cin >> rubbish;
            cin >> first;
            cin >> rubbish;
            cin >> value;
            ek.AddEdge(s,first,value);//超级源点与供电点相连
        }
        for (int i = 0; i < nc; i++)
        {
            cin >> rubbish;
            cin >> first;
            cin >> rubbish;
            cin >> value;
            ek.AddEdge(first,t,value);;//超级汇点与消费点相连
        }
        cout<<ek.Maxflow(s,t)<<endl;
    }
    return 0;
}