网络流24题——餐巾计划问题(有源汇的最小费用可行流)

链接：https://www.oj.swust.edu.cn/oj/problem/show/1745

分析：

       我的想法是，首先每天的餐巾数有限制，所以先拆点，从Xi向Yi连一条容量和下界均为ri的边，Xi看作每天得到的新餐巾（洗好的+新买的），从s向Xi连一条容量为inf，费用为p的边；再从Xi向X(i+m)连一条容量为inf，费用为f的边，向X(i+n)连容量为inf，费用为s的边，还有不洗的那些，直接留到下一天，就是从Xi到X(i+1)连一条容量为inf，费用为0的边(最后一天的直接连到t上)，然后求这个图的最小费用可行流。

       带下界的最小费用可行流求法：先从t到s连容量为inf，费用为0的边，变成无源无汇的图，增加附加源汇S、T，然后对于每一条从u到v下界为b上界为c费用为w的边，拆成u到T容量为b、费用为w的边和S到v容量为b费用为0的边，再把u到v的边容量改为c-b，求S-T最小费用最大流即为所求，当且仅当附加边均满流时问题有解。

        题解是二分图。。比我的做法简单。。

我的做法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=5e3+5,INF=1e9;
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,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    int inq[maxn],d[maxn],p[maxn],a[maxn];
    ll flow,cost;
    void init(int n){
        this->n=n;
        flow=0,cost=0;
        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));
        G[from].push_back(edges.size()-2);
        G[to].push_back(edges.size()-1);
    }

    bool spfa(int s,int t){
        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+=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;
    }
    ll Mincost(int s,int t){
        while(spfa(s,t));
        return cost;
    }
}mcmf;

int N,p,m,f,n,s,r[maxn];
void add(int k,int r){
    mcmf.AddEdge(0,k,INF,p);
    mcmf.AddEdge(2*N+2,k+N,r,0);
    mcmf.AddEdge(k,2*N+3,r,0);
    if(k+1<=N)mcmf.AddEdge(k+N,k+1+N,INF,0);
    if(k+m<=N)mcmf.AddEdge(k+N,k+m,INF,f);
    if(k+n<=N)mcmf.AddEdge(k+N,k+n,INF,s);
}
int main(){
//    freopen("e:\in.txt","r",stdin);
    scanf("%d%d%d%d%d%d",&N,&p,&m,&f,&n,&s);
    for(int i=1;i<=N;i++)scanf("%d",&r[i]);
    mcmf.init(2*N+4);
    mcmf.AddEdge(2*N,2*N+1,INF,0);
    mcmf.AddEdge(2*N+1,0,INF,0);
    for(int i=1;i<=N;i++){
        add(i,r[i]);
    }
    printf("%lld
",mcmf.Mincost(2*N+2,2*N+3));
    return 0;
}

二分图做法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=5e3+5,INF=1e9;
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,s,t;
    ll flow,cost;
    vector<Edge> edges;
    vector<int> G[maxn];
    int inq[maxn],d[maxn],p[maxn],a[maxn];

    void init(int n){
        flow=0,cost=0;
        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));
        G[from].push_back(edges.size()-2);
        G[to].push_back(edges.size()-1);
    }

    bool spfa(int s,int t,ll &flow,ll &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+=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;
    }
    ll Mincost(int s,int t){
        while(spfa(s,t,flow,cost));
        return cost;
    }
}isap;

int N,p,m,f,n,s,r[maxn];
void add(int k,int r){
    isap.AddEdge(0,k,r,0);
    isap.AddEdge(0,k+N,INF,p);
    isap.AddEdge(k+N,2*N+1,r,0);
    if(k+1<=N)isap.AddEdge(k,k+1,INF,0);
    if(k+m<=N)isap.AddEdge(k,k+m+N,INF,f);
    if(k+n<=N)isap.AddEdge(k,k+n+N,INF,s);
}
int main(){
//    freopen("e:\in.txt","r",stdin);
    scanf("%d%d%d%d%d%d",&N,&p,&m,&f,&n,&s);
    isap.init(2*N+2);
    int r;
    for(int i=1;i<=N;i++){
        scanf("%d",&r);
        add(i,r);
    }
    printf("%lld
",isap.Mincost(0,2*N+1));
    return 0;
}