标签：like   res   lap   any   and   无向图   connect   pop   route   

链接:

https://vjudge.net/problem/HDU-4280

题意:

In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships.
　　You have a transportation company there. Some routes are opened for passengers. Each route is a straight line connecting two different islands, and it is bidirectional. Within an hour, a route can transport a certain number of passengers in one direction. For safety, no two routes are cross or overlap and no routes will pass an island except the departing island and the arriving island. Each island can be treated as a point on the XY plane coordinate system. X coordinate increase from west to east, and Y coordinate increase from south to north.
　　The transport capacity is important to you. Suppose many passengers depart from the westernmost island and would like to arrive at the easternmost island, the maximum number of passengers arrive at the latter within every hour is the transport capacity. Please calculate it.

思路:

考虑无向图,只需给反向边加上容量,同时图的点很多,使用ISAP算法.
找了很久,看了很久..找的代码bfs写错了(貌似?).最后很久才改出来.

代码:

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1e5+10;
const int INF = 1e9;

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

int Pre[MAXN], Cur[MAXN];
int Num[MAXN], Vis[MAXN];
int Dis[MAXN];
vector<int> G[MAXN];
vector<Edge> edges;
int n, m, s, t;

void AddEdge(int from, int to, int cap)
{
    edges.push_back(Edge{from, to, 0, cap});
    edges.push_back(Edge{to, from, 0, cap});
    int len = edges.size();
    G[from].push_back(len - 2);
    G[to].push_back(len - 1);
}

void Bfs()
{
    memset(Vis, 0, sizeof(Vis));
    queue<int> que;
    que.push(t);
    Vis[t] = 1;
    Dis[t] = 0;
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
        for (int i = 0;i < G[u].size();i++)
        {
            Edge &e = edges[G[u][i]^1];
            if (Vis[e.from] == 0 && e.cap > e.flow)
            {
                Vis[e.from] = 1;
                que.push(e.from);
                Dis[e.from] = Dis[u]+1;
            }
        }
    }
}

int Augment()
{
//    cout << 1 << endl;
    int x = t, flow = INF;
    while (x != s)
    {
        Edge &e = edges[Pre[x]];
//        cout << e.from << ' ' << e.to << endl;
        flow = min(flow, e.cap-e.flow);
        x = e.from;
    }
//    cout << flow << endl;
    x = t;
    while (x != s)
    {
        edges[Pre[x]].flow += flow;
        edges[Pre[x]^1].flow -= flow;
        x = edges[Pre[x]].from;
    }
    return flow;
}

int MaxFlow()
{
    int flow = 0;
    Bfs();
    memset(Num, 0, sizeof(Num));
    for (int i = 0;i < n;i++)
        Num[Dis[i]]++;
    int x = s;
    memset(Cur, 0, sizeof(Cur));
    while (Dis[s] < n)
    {
        if (x == t)
        {
            flow += Augment();
            x = s;
        }
        bool ok = false;
        for (int i = Cur[x];i < G[x].size();i++)
        {
            Edge &e = edges[G[x][i]];
            if (e.cap > e.flow && Dis[x] == Dis[e.to]+1)
            {
                ok = true;
                Pre[e.to] = G[x][i];
                Cur[x] = i;
                x = e.to;
                break;
            }
        }
        if (!ok)
        {
            int line = n-1;
            for (int i = 0;i < G[x].size();i++)
            {
                Edge &e = edges[G[x][i]];
                if (e.cap > e.flow)
                    line = min(line, Dis[e.to]);
            }
            if (--Num[Dis[x]] == 0)
                break;
            Dis[x] = line+1;
            Num[Dis[x]]++;
            Cur[x] = 0;
            if (x != s)
                x = edges[Pre[x]].from;
        }
    }
    return flow;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d %d", &n, &m);
        for (int i = 1;i <= n;i++)
            G[i].clear();
        edges.clear();
        int mmin = INF, mmax = -INF;
        int u, v, w;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d %d", &u, &v);
            if (u < mmin)
                s = i, mmin = u;
            if (u > mmax)
                t = i, mmax = u;
        }
        for (int i = 1;i <= m;i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            AddEdge(u, v, w);
        }
        int res = MaxFlow();
        printf("%d\n", res);
    }

    return 0;
}

HDU-4280-Island Transport(网络流,最大流, ISAP)

标签：like   res   lap   any   and   无向图   connect   pop   route   

原文地址：https://www.cnblogs.com/YDDDD/p/11329074.html