• HDU 5889 Barricade


    最短路,最小割,网络流。

    可以根据$dis[u]+1$与$dis[v]$的大小关系判断$<u,v>$是否为最短路上的边,可以处理出一个只包含最短路的$DAG$,然后求这个$DAG$的最小割就可以了。

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<vector>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<iostream>
    using namespace std;
    typedef long long LL;
    const double pi=acos(-1.0),eps=1e-6;
    void File()
    {
        freopen("D:\in.txt","r",stdin);
        freopen("D:\out.txt","w",stdout);
    }
    template <class T>
    inline void read(T &x)
    {
        char c=getchar(); x=0;
        while(!isdigit(c)) c=getchar();
        while(isdigit(c)) {x=x*10+c-'0'; c=getchar();}
    }
    
    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;
    }
    
    int h[maxn],sz,T;
    struct X
    {
        int u,v,w,nx;
    }ee[maxn];
    int dis[maxn],flag[maxn];
    
    void add(int a,int b,int c)
    {
        ee[sz].u=a; ee[sz].v=b; ee[sz].w=c;
        ee[sz].nx=h[a]; h[a]=sz++; 
    }
    
    void spfa()
    {
        for(int i=1;i<=n;i++) dis[i]=INF ,flag[i]=0;
        dis[1]=0; queue<int>Q; Q.push(1); flag[1]=1;
    
        while(!Q.empty())
        {
            int top=Q.front(); Q.pop(); flag[top]=0;
            for(int i=h[top];i!=-1;i=ee[i].nx)
            {
                if(dis[top]+1<dis[ee[i].v])
                {
                    dis[ee[i].v]=dis[top]+1;
                    if(flag[ee[i].v]==0)
                    {
                        flag[ee[i].v]=1;
                        Q.push(ee[i].v);
                    }
                }
            }
        }
    }
    
    int main()
    {
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&n,&m); sz=0;
            memset(h,-1,sizeof h);
            for(int i=1;i<=m;i++)
            {
                int u,v,w; scanf("%d%d%d",&u,&v,&w);
                add(u,v,w); add(v,u,w);
            }
            spfa();
    
            init();
            for(int i=0;i<sz;i++)
            {
                if(dis[ee[i].u]+1==dis[ee[i].v])
                {
                    AddEdge(ee[i].u,ee[i].v,ee[i].w);
                }
            }
            s=1; t=n;
            printf("%d
    ",dinic(s,t));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5887018.html
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