hdu 4494 Teamwork （可行流的最小流）

标签：图论

             去年通话邀请赛的B题，当时竟然过的那么少。。。明明是一道很裸的可行流最小流麽。。只要对每种人分别求一下可行最小流加起来就可以了。建图是对每个点拆点，容量上下届都设为v[i]，然后每个点间能连边的直接连边就可以了。然后在这个图的基础上转化为可行流最小流，求一下就可以了。。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define inf 0x3f3f3f3f
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;

const int maxn = 440;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, cap, flow;
    Edge() {}
    Edge(int from, int to, int cap, int flow)
        :from(from), to(to), cap(cap), flow(flow) {}
};

struct ISAP
{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
    bool vis[maxn];        // BFS使用
    int d[maxn];           // 从起点到i的距离
    int cur[maxn];        // 当前弧指针
    int p[maxn];          // 可增广路上的上一条弧
    int num[maxn];        // 距离标号计数

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

    bool BFS()
    {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(t);
        vis[t] = 1;
        d[t] = 0;
        while(!Q.empty())
        {
            int x = Q.front();
            Q.pop();
            for(int i = 0; i < G[x].size(); i++)
            {
                Edge& e = edges[G[x][i]^1];
                if(!vis[e.from] && e.cap > e.flow)
                {
                    vis[e.from] = 1;
                    d[e.from] = d[x] + 1;
                    Q.push(e.from);
                }
            }
        }
        return vis[s];
    }

    void init(int n)
    {
        this->n = n;
        for(int i = 0; i < n; i++) G[i].clear();
        edges.clear();
    }

    int Augment()
    {
        int x = t, a = INF;
        while(x != s)
        {
            Edge& e = edges[p[x]];
            a = min(a, e.cap-e.flow);
            x = edges[p[x]].from;
        }
        x = t;
        while(x != s)
        {
            edges[p[x]].flow += a;
            edges[p[x]^1].flow -= a;
            x = edges[p[x]].from;
        }
        return a;
    }

    int Maxflow(int s, int t, int need)
    {
        this->s = s;
        this->t = t;
        int flow = 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)
            {
                flow += Augment();
                if(flow >= need) return flow;
                x = s;
            }
            int ok = 0;
            for(int i = cur[x]; i < G[x].size(); i++)
            {
                Edge& e = edges[G[x][i]];
                if(e.cap > e.flow && d[x] == d[e.to] + 1)   // Advance
                {
                    ok = 1;
                    p[e.to] = G[x][i];
                    cur[x] = i; // 注意
                    x = e.to;
                    break;
                }
            }
            if(!ok)   // Retreat
            {
                int m = n-1; // 初值注意
                for(int i = 0; i < G[x].size(); i++)
                {
                    Edge& e = edges[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; // 注意
                if(x != s) x = edges[p[x]].from;
            }
        }
        return flow;
    }
} sol;

int n, m;

struct Point
{
    int x, y, b, e;
    int v[7];
    void inpt()
    {
        scanf("%d%d%d%d", &x, &y, &b, &e);
        e += b;
        for(int i = 0; i < m; i ++)
            scanf("%d", &v[i]);
    }
}p[maxn];

bool ok(Point a, Point b)
{
    int len = (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
    return b.b > a.e && len <= (b.b - a.e) * (b.b - a.e);
}

int solve(int idx)
{
    sol.init(n * 2 + 5);
    int S = 0, T = 2 * n + 1, SS = 2 * n + 2, ST = SS + 1;
    for(int i = 1; i <= n; i ++)
    {
//        sol.AddEdge(i, i + n, 0);
        sol.AddEdge(S, i, INF);
        sol.AddEdge(i + n, T, INF);

        sol.AddEdge(SS, i + n, p[i].v[idx]);
        sol.AddEdge(i, ST, p[i].v[idx]);
    }
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= n; j ++)
        {
            if(!ok(p[i], p[j])) continue;
            sol.AddEdge(i + n, j, p[i].v[idx]);
        }
    sol.Maxflow(SS, ST, INF);
    sol.AddEdge(T, S, INF);
    sol.Maxflow(SS, ST, INF);
    return sol.edges[sol.edges.size() - 2].flow;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T --)
    {
        scanf("%d%d", &n, &m);
        n --;
        scanf("%d%d", &p[0].x, &p[0].y);
        p[0].b = p[0].e = 0;
        for(int i = 1; i <= n; i ++)
            p[i].inpt();
        int ans = 0;
        for(int i = 0; i < m; i ++)
            ans += solve(i);
        printf("%d\n", ans);
    }
}

hdu 4494 Teamwork （可行流的最小流）,布布扣,bubuko.com

hdu 4494 Teamwork （可行流的最小流）

标签：图论

原文地址：http://blog.csdn.net/ok_again/article/details/24872897