Optimal Bus Route Design UVA - 1349

问题

分析

按照紫书上的解释，将一个点拆分成两个点X_i,Y_i，就变成了一个二分图匹配问题，方法和上一题UVA1658优点相似，将入弧和出弧分别连接在两个点X_i,Y_i上，两点间不连接，因为他们实际上是一个点，不用计入cost中，然后添加源点宿点，源点连接在X_i上，Y_i连接到宿点上，求出源点，宿点之间的最小费用最大流，然后将二部图最终的对应点连接起来，就是最小权和的圈

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=205,Inf=0x3f3f3f3f;

struct Edge{
    int from,to,cap,flow,cost;
    Edge(int from,int to,int cap,int flow,int cost): from(from),to(to),cap(cap),flow(flow),cost(cost){}
};

struct MCMF{
    int n,m;
    vector<Edge> edges;
    vector<int> G[maxn];
    int a[maxn];
    int p[maxn];
    int d[maxn];
    int inq[maxn];

    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,int 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; a[s]=Inf; p[s]=0;
        queue<int> Q;
        Q.push(s);
        while(!Q.empty()){
            int x=Q.front();
            Q.pop();
            inq[x]=0;
            for(int i=0;i<G[x].size();++i){
                Edge &e=edges[G[x][i]];
                if(d[e.to]>d[x]+e.cost && e.cap>e.flow){
                    d[e.to]=d[x]+e.cost;
                    a[e.to]=min(a[x],e.cap-e.flow);
                    p[e.to]=G[x][i];
                    if(!inq[e.to]){
                        inq[e.to]=1;
                        Q.push(e.to);
                    }
                }
            }
        }
        if(d[t]==Inf) return false;
        flow+=a[t];
        cost+=(long long)a[t]*d[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;
    }
};

MCMF mcmf;
int n,a,b;
int main(void){
    while(cin>>n && n){
        mcmf.init(2*n+2);
        for(int i=1;i<=n;++i){
            mcmf.AddEdge(0,i,1,0);  //源点到X
            mcmf.AddEdge(i+n,(n<<1)+1,1,0);  //Y到宿点
        }
        for(int i=1;i<=n;++i){
            while(scanf("%d",&a)==1 && a){
                scanf("%d",&b);
                mcmf.AddEdge(i,a+n,1,b);
            }
        }
        long long cost=0;
        int flow=mcmf.MincostMaxflow(0,2*n+1,cost);
        if(flow==n) printf("%lld\n",cost);
        else printf("N\n");
    }
    return 0;
}