• poj3422 拆点法x->x'建立两条边+最小费用最大流


    /**
    题目:poj3422 拆点法+最小费用最大流
    链接:http://poj.org/problem?id=3422
    题意:给定n*n的矩阵,含有元素值,初始sum=0.每次从最左上角开始出发,每次向右或者向下一格。终点是右下角。
    每经过一个格子,获取它的值,并把该格子的值变成0.问经过k次从左上角到右下角。能得到的数值和最大多少。
    
    思路:我觉得本题元素值全是非负数。要不然不可以过。很多网上的博客代码在有负数情况下过不了。
    拆点法+最小费用最大流
    
    建图:
    每一个格子x,拆成x,xi, x向xi连两条边,其一:x->xi,cap=1,cost=-wx;其二:x->xi,cap=k-1,cost=0;表示x这个格子可以经过k次,
    第一次获得值为wx,之后经过它只能获得0.
    
    左上角格子x,   s->x,cap=k,cost=0;
    右下角格子x,   xi->t,cap=k,coste=0;
    
    如果x的右边的格子或者下面的格子是y,  xi->y,cap=k,cost=0;
    
    
    
    */
    #include<iostream>
    #include<cstring>
    #include<vector>
    #include<map>
    #include<cstdio>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    using namespace std;
    typedef long long LL;
    const int INF = 0x3f3f3f3f;
    const int N = 5500;
    struct Edge{
        int from, to, cap, flow, cost;
        Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
    };
    struct MCMF{
        int n, m;
        vector<Edge> edges;
        vector<int> G[N];
        int inq[N];
        int d[N];
        int p[N];
        int a[N];
    
        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,long long cost){
            edges.push_back(Edge(from,to,cap,0,cost));
            edges.push_back(Edge(to,from,0,0,-cost));
            m = edges.size();
            G[from].push_back(m-2);
            G[to].push_back(m-1);
        }
    
        bool BellmanFord(int s,int t,int &flow,long long &cost){
            for(int i = 0; i <= n; i++) d[i] = INF;
            memset(inq, 0, sizeof inq);
            d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
    
            queue<int>  Q;
            Q.push(s);
            while(!Q.empty()){
                int u = Q.front(); Q.pop();
                inq[u] = 0;
                for(int i = 0; i < G[u].size(); i++){
                    Edge& e = edges[G[u][i]];
                    if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
                        d[e.to] = d[u]+e.cost;
                        p[e.to] = G[u][i];
                        a[e.to] = min(a[u],e.cap-e.flow);
                        if(!inq[e.to]) {Q.push(e.to); inq[e.to] = 1;}
                    }
                }
            }
            if(d[t]==INF) return false;
            flow += a[t];
            cost += (long long)d[t]*(long long)a[t];
            for(int u = t; u!=s; u = edges[p[u]].from){
                edges[p[u]].flow+=a[t];
                edges[p[u]^1].flow-=a[t];
            }
            return true;
        }
        int MincostMaxflow(int s,int t,long long &cost){
            int flow = 0;
            cost = 0;
            while(BellmanFord(s,t,flow,cost));
            return flow;
        }
    };
    int main()
    {
        int n, k;
        while(scanf("%d%d",&n,&k)==2)
        {
            int w, s = 0, t = n*n+1;
            MCMF mcmf;
            mcmf.init(t*2);
            mcmf.AddEdge(s,1,k,0);
            for(int i = 1; i <= n; i++){
                for(int j = 1;j <= n; j++){
                    scanf("%d",&w);
                    int x = (i-1)*n+j, y = x+t;
                    mcmf.AddEdge(x,y,1,-w);
                    mcmf.AddEdge(x,y,k-1,0);
                    if(i==n&&j==n){
                        mcmf.AddEdge(y,t,k,0);
                    }
                    if(j+1<=n){
                        int m = (i-1)*n+j+1;
                        mcmf.AddEdge(y,m,k,0);
                    }
                    if(i+1<=n){
                        int m = i*n+j;
                        mcmf.AddEdge(y,m,k,0);
                    }
                }
            }
            LL cost;
            mcmf.MincostMaxflow(s,t,cost);
            printf("%lld
    ",-cost);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/xiaochaoqun/p/7224591.html
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