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  • HDU 3338 Kakuro Extension

    网络最大流

    TLE了两天的题目。80次Submit才AC,发现是刘汝佳白书的Dinic代码还可以优化。。。。。瞬间无语。。。。。

    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<vector>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    const int maxn = 30000 + 10;
    const int INF = 0x7FFFFFFF;
    struct Edge
    {
        int from, to, cap, flow;
        Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f) {}
    };
    vector<Edge>edges;
    vector<int>G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    int n, m, s, t;
    
    void init()
    {
        for (int i = 0; i < maxn; 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));
        int w = edges.size();
        G[from].push_back(w - 2);
        G[to].push_back(w - 1);
    }
    bool BFS()
    {
        memset(vis, 0, sizeof(vis));
        queue<int>Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;
        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)
            {
                edges[G[x][i]].flow+=f;
                edges[G[x][i] ^ 1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0) break;
            }
        }
        if(!flow) d[x] = -1;
        return flow;
    }
    int dinic(int s, int t)
    {
        int flow = 0;
        while (BFS())
        {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }
    
    //输入输出
    int N, M;
    char S[1000];
    int Map1[105][105];//横向
    int Map2[105][105];//纵向
    int BH[105][105];
    int FLAG[maxn];
    
    int main()
    {
        while (~scanf("%d%d", &N, &M))
        {
    
            edges.clear();
            for (int i = 0; i<maxn; i++) G[i].clear();
            memset(Map1, 0, sizeof(Map1));
            memset(Map2, 0, sizeof(Map2));
            memset(BH, 0, sizeof(BH));
            for (int i = 1; i <= N; i++)
                for (int j = 1; j <= M; j++)
                {
                    scanf("%s", S);
                    if (S[3] == '.') continue;
                    if (S[3] == 'X') Map1[i][j] = -1, Map2[i][j] = -1;
                    else
                    {
                        if (S[2] == 'X')
                        {
                            Map2[i][j] = -1;
                            Map1[i][j] = (S[4] - '0') * 100 + (S[5] - '0') * 10 + (S[6] - '0') * 1;
                        }
                        else if (S[4] == 'X')
                        {
                            Map1[i][j] = -1;
                            Map2[i][j] = (S[0] - '0') * 100 + (S[1] - '0') * 10 + (S[2] - '0') * 1;
                        }
                        else
                        {
                            Map1[i][j] = (S[4] - '0') * 100 + (S[5] - '0') * 10 + (S[6] - '0') * 1;
                            Map2[i][j] = (S[0] - '0') * 100 + (S[1] - '0') * 10 + (S[2] - '0') * 1;
                        }
                    }
                }
            int Tot = 1;
            for (int i = 1; i <= N; i++)
                for (int j = 1; j <= M; j++)
                    if (Map1[i][j] == 0)
                        BH[i][j] = Tot, Tot++;
            int Duan = Tot - 1;//方格编号1--Duan
            s = 0;
            t = 30000;
            for (int i = 1; i <= N; i++)
                for (int j = 1; j <= M; j++)
                {
                    if (Map1[i][j] != -1 && Map1[i][j] != 0)
                    {
                        int Zong = 0;
                        for (int k = j + 1;; k++)
                        {
                            if (k>M || Map1[i][k] != 0) break;
                            AddEdge(Tot, BH[i][k], 8);
                            Zong++;
                        }
                        AddEdge(s, Tot, Map1[i][j] - Zong);
                        Tot++;
                    }
                }
            for (int i = 1; i <= N; i++)
                for (int j = 1; j <= M; j++)
                {
                    if (Map2[i][j] != -1 && Map2[i][j] != 0)
                    {
                        int Zong = 0;
                        for (int k = i + 1;; k++)
                        {
                            if (Map2[k][j] != 0 || k>N) break;
                            AddEdge(BH[k][j], Tot, 8);
                            Zong++;
                        }
                        AddEdge(Tot, t, Map2[i][j] - Zong);
                        Tot++;
                    }
                }
            dinic(s, t);
            memset(FLAG, 0, sizeof FLAG);
            for (int i = 0; i<edges.size(); i = i + 2)
                if (edges[i].from >= 1 && edges[i].from <= Duan)
                    FLAG[edges[i].from] = edges[i].flow;
            for (int i = 1; i <= N; i++)
            {
                for (int j = 1; j <= M; j++)
                {
                    if (BH[i][j] == 0) printf("_");
                    else printf("%d", FLAG[BH[i][j]] + 1);
                    if (j<M) printf(" ");
                }
                printf("
    ");
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/4757238.html
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