CCF (Duct Cleaning): Minimum Cost Maximum

Duct Cleaning

201812-5

• The lower bound need to clean the pipe 1,
• Does not require cleaning pipe lower bound 0,
• After re-pipe the upper bound is positive infinity,
• After unrepeatable pipe 1 is the upper bound.
• It belongs to the passive sinks have no capacity lower bound of the minimum cost feasible flow. The solution is to first add a source and a sink, with a lower limit of the arc and to transform into capacity per a lower limit of 0 and an upper limit of an arc cb, add two point to point source from the point x the upper limit of the arc b, y and directed from the source point of the arc b for the upper limit.
• Finally, the minimum cost flow st after only requires the transformation of the line. However, if and only if the additional full arc when the original network feasible flow.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
using namespace std;
const int N=210;
const int INF=0X3F3F3F3F; 
int n,m,e;
int no[N];
int overload;//需要清扫的道路数目
int sum;//根据需要清扫的道路数目可知必须要消耗的包子数目,注意这里仅仅是需要清扫的道路的必须的包子数目，而不是全部的
struct Edge {
    int from, to, cap, flow, cost;
};
struct MCMF {
    int n, m;
    vector<Edge> edges;
    vector<int> G[N];
    int d[N], inq[N], p[N], 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, 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 spfa(int s, int t, int &flow, int &cost) {
        //M(inq, 0); M(d, INF);
        memset(inq,0,sizeof(inq));
        memset(d,INF,sizeof(d));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
        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;
                    p[e.to] = G[x][i];
                    a[e.to] = min(a[x], e.cap-e.flow);
                    if (inq[e.to]) continue;
                    q.push(e.to); inq[e.to] = 1;
                }
            }
        }
        if (d[t] == INF) return false;
        flow += a[t];
        cost += d[t] * a[t];
        int u = t;
        while (u != s) {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }

    int Mincost(int s, int t) {
        int flow = 0, cost = 0;
        while (spfa(s, t, flow, cost));
        if(flow!=overload)
            return -1;
        return cost;
    }

}solver;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t,s;
    cin>>t>>s>>e;
    while(t--){
        overload=0;
        sum=0;
        int st,ed;
        cin>>n>>m;
        st=0,ed=n+1;
        memset(no,0,sizeof(no));
        solver.init(n+1);
        for(int i=0;i<m;i++){
            int x,y;
            char c;
            cin>>x>>y>>c;
            if(c=='A'){//需要被清理，可以走无数遍，上限为无穷，下限为1
                solver.AddEdge(x,y,INF,e);
                no[x]--;
                no[y]++;
                sum+=e;
            }else if(c=='B'){//需要被清理，只能走一遍，上限为1，下限也为1
                no[x]--;
                no[y]++;
                sum+=e;
            }else if(c=='C'){//不要被清理，可以走无数遍，上限为无穷，下限为0
                solver.AddEdge(x,y,INF,e);
            }else{//不需要被清理，但是只能走一遍，上限为1，下限为0
                solver.AddEdge(x,y,1,e);
            }
        }
        for(int i=1;i<=n;i++){//因为这里的下限刚好是1，所以可以采用++，--的方法来统计一个顶点共有多少次上限-下限
            if(no[i]>0){//后结点
                solver.AddEdge(st,i,no[i],0);//一下两条路属于附加结点，附加边，所以费用是0
                overload+=no[i];
            }else if(no[i]<0){//前结点
                solver.AddEdge(i,ed,-no[i],0);
            }
        }
        cout<<solver.Mincost(st,ed)+sum<<endl;
    }
    //system("pause");
    return 0;
}