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  • POJ 3469 Dual Core CPU 最大流

    划分成两个集合使费用最小,可以转成最小割,既最大流。

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<iostream>
    #include<sstream>
    #include<cmath>
    #include<climits>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    #define pb(a) push(a)
    #define INF 0x1f1f1f1f
    #define lson idx<<1,l,mid
    #define rson idx<<1|1,mid+1,r
    #define PI  3.1415926535898
    template<class T> T min(const T& a,const T& b,const T& c)
    {
        return min(min(a,b),min(a,c));
    }
    template<class T> T max(const T& a,const T& b,const T& c)
    {
        return max(max(a,b),max(a,c));
    }
    void debug()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("data.in","r",stdin);
        // freopen("d:\out1.txt","w",stdout);
    #endif
    }
    int getch()
    {
        int ch;
        while((ch=getchar())!=EOF)
        {
            if(ch!=' '&&ch!='
    ')return ch;
        }
        return EOF;
    }
    
    struct Edge
    {
        int to,cap;
    };
    const int maxn = 20010;
    vector<int> g[maxn];
    vector<Edge> edge;
    int n,m,s,t;
    int d[maxn];
    int vis[maxn];
    int cur[maxn];
    
    void add(int u,int v,int cap)
    {
        //printf("%d -> %d : %d
    ", u, v, cap);
        edge.push_back((Edge){v, cap});
        g[u].push_back(edge.size() - 1);
        edge.push_back((Edge){u, 0});
        g[v].push_back(edge.size() - 1);
    }
    void build()
    {
        for(int i = 1; i <= n; i++)
            g[i].clear();
        edge.clear();
        s = n + 1;
        t = s + 1;
        for(int i = 1; i <= n; i++)
        {
            int a,b;
            scanf("%d%d", &a, &b);
            add(s, i, a);
            add(i, t, b);
        }
        for(int i = 1; i <= m; i++)
        {
            int a,b,w;
            scanf("%d%d%d", &a, &b, &w);
            add(a, b, w);
            add(b, a, w);
        }
    }
    
    bool bfs()
    {
        memset(vis, 0, sizeof(vis));
        d[s] = 0;
        queue<int> q;
        q.push(s);
        vis[s] = true;
        while(!q.empty())
        {
            int x = q.front(); q.pop();
            for (int i = 0; i < g[x].size(); i++)
            {
                Edge &e = edge[g[x][i]];
                if(e.cap>0 && !vis[e.to])
                {
                    vis[e.to] = true;
                    q.push(e.to);
                    d[e.to] = d[x] + 1;
                }
            }
        }
        return vis[t];
    }
    
    int dfs(int u,int f)
    {
        if(u==t || f==0) return f;
        for(int &i = cur[u]; i < g[u].size(); i++)
        {
            Edge &e = edge[g[u][i]];
            if(e.cap>0 && d[e.to] == d[u] + 1)
            {
                int d = dfs(e.to, min(f,e.cap));
                if(d>0)
                {
                    e.cap -= d;
                    edge[g[u][i]^1].cap += d;
                    return d;
                }
            }
        }
        return 0;
    }
    
    int max_flow()
    {
        int res = 0;
        while(bfs())
        {
            memset(cur, 0, sizeof(cur));
            int d;
            while(d=dfs(s,INF))
            {
                res += d;
            }
        }
        return res;
    }
    int main()
    {
        debug();
        while(scanf("%d%d", &n, &m) != EOF)
        {
            build();
            printf("%d
    ", max_flow());
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/BMan/p/3713704.html
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