hunnu OJ 11567 Escaping(拆点型最大匹配/网络流)

标签：

题意：

思路：建立源点跟汇点，将存在需要救援的房间与源点相连，容量为该房间的人数。将存在救生设备的房间与汇点相连，容量为救生设备的数目。中间点则考虑在t分钟内是否能够到达进行连边即可。

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
const int maxn = 10000 + 50;
const int INF = 1e9;

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

struct Dinic {
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void init(int n)
    {
        this->n = 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
        });
        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(s);
        vis[s] = 1;
        d[s] = 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]];
                if(!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x, int a)
    {
        if(x==t || a==0) return a;
        int flow = 0, f;
        for(int& i=cur[x]; i<G[x].size(); ++i) {
            Edge& e = edges[G[x][i]];
            if(d[x]+1==d[e.to] && (f=DFS(e.to, min(a, e.cap-e.flow)))>0) {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if(a==0) break;
            }
        }
        return flow;
    }

    int MaxFlow(int s, int t)
    {
        this->s = s;
        this->t = t;
        int flow = 0;
        while(BFS()) {
            memset(cur, 0, sizeof cur );
            flow += DFS(s, INF);
        }
        return flow;
    }
};

Dinic solve;

int mtx[50][50];

int dist(int x1, int y1, int x2, int y2)
{
    return abs(x1-x2) + abs(y1-y2);
}

int main()
{
#ifdef LOCAL_DEFINE
    freopen("in.cpp", "r", stdin);
    freopen("out.cpp", "w", stdout);
#endif // LOCAL_DEFINE
    int n, tt, x;
    int s, t;
    while(~scanf("%d%d", &n, &tt)) {
        int lmt = n*n;
        s = lmt*2 + 1;
        t = lmt*2 + 2;

        solve.init(t+2);

        for(int i=0; i<n; ++i) {
            for(int j=0; j<n; ++j) {
                scanf("%d", &mtx[i][j]);
                if(mtx[i][j]>0) {
                    solve.AddEdge( s, n*i+j, mtx[i][j]);
                }
            }
        }

        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++) {
                scanf("%d",&x);
                if(x>0) {
                    int u = n*i+j + lmt;
                    solve.AddEdge(u, t, x);

                    for(int i1=0; i1<n; ++i1)
                        for(int j1=0; j1<n; ++j1) if(mtx[i1][j1]>0 && dist(i1, j1, i, j)<=tt){
                        int v = n*i1 + j1;
                        solve.AddEdge(v, u, INF);
                    }
                }
            }
        int ans = solve.MaxFlow(s, t);
        printf("%d\n", ans);
    }
    return 0;
}

hunnu OJ 11567 Escaping(拆点型最大匹配/网络流)

标签：

原文地址：http://blog.csdn.net/u012313382/article/details/47399475