• UVA 10480 Sabotage


    最小割+输出方案

    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<vector>
    #include<queue>
    #include<algorithm>
    using namespace std;
    
    const int maxn=1000+10;
    const int INF=0x7FFFFFFF;
    
    struct Edge
    {
        int from,to,cap,flow;
    };
    vector<Edge>edges;
    vector<int>G[maxn];
    vector<int>CG[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    int n,m,s,t;
    Edge P[maxn];
    
    //求出层次网络
    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];
    }
    
    
    //加边
    void AddEdge(int from,int to,int cap)
    {
        Edge r;
        r.from=from;
        r.to=to;
        r.cap=cap;
        r.flow=0;
        edges.push_back(r);
        Edge d;
        d.from=to;
        d.to=from;
        d.cap=0;
        d.flow=0;
        edges.push_back(d);
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    
    //每个阶段来一次DFS增广
    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;
    }
    
    //多个阶段,多次建立层次网络。
    int Maxflow(int ss,int tt)
    {
        int flow=0;
        while(BFS())
        {
            memset(cur,0,sizeof(cur));
            flow+=DFS(ss,INF);
        }
        return flow;
    }
    
    int N,M,Cost;
    int U[maxn],V[maxn];
    int FF[maxn];
    
    void dfs(int x)
    {
        FF[x]=1;
        for(int i=0; i<CG[x].size(); i++)
            if(!FF[CG[x][i]])
                dfs(CG[x][i]);
    }
    
    int main()
    {
        while(~scanf("%d%d",&N,&M))
        {
            if(!N&&!M) break;
            edges.clear();
            for(int i=0; i<maxn; i++) G[i].clear();
            for(int i=0; i<maxn; i++) CG[i].clear();
            memset(FF,0,sizeof FF);
            s=1;
            t=2;
            for(int i=1; i<=M; i++)
            {
                scanf("%d%d%d",&U[i],&V[i],&Cost);
                AddEdge(U[i],V[i],Cost);
                AddEdge(V[i],U[i],Cost);
            }
            Maxflow(s,t);
            int Tot=0;
            for(int i=0; i<edges.size(); i=i+2)
            {
                P[Tot].from=edges[i].from;
                P[Tot].to=edges[i].to;
                P[Tot].cap=edges[i].cap;
                P[Tot].flow=edges[i].flow;
                Tot++;
            }
            for(int i=0; i<Tot; i++)
            {
                if(P[i].cap-P[i].flow>0)
                    CG[P[i].from].push_back(P[i].to);
                if(P[i].flow>0)
                    CG[P[i].to].push_back(P[i].from);
            }
            dfs(s);
            for(int i=1; i<=M; i++)
                if((FF[U[i]]==0&&FF[V[i]]==1)||(FF[V[i]]==0&&FF[U[i]]==1))
                    printf("%d %d
    ",U[i],V[i]);
            printf("
    ");
        }
        return 0;
    }
  • 相关阅读:
    [leetcode-79-Word Search]
    [leetcode-563-Binary Tree Tilt]
    [leetcode-561-Array Partition I]
    [leetcode-556-Next Greater Element III]
    [leetcode-554-Brick Wall]
    [leetcode-557-Reverse Words in a String III]
    [leetcode-532-K-diff Pairs in an Array]
    dev 官网
    DataTable 设置primarykey 后进行 Merge操作
    对称矩阵 一个简单的小把戏
  • 原文地址:https://www.cnblogs.com/zufezzt/p/4748230.html
Copyright © 2020-2023  润新知