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第一问裸的最大流。

第二问给每一条边加上容量无限,费用为输入的边,新建起点终点,

新起点连1,费用为0,容量为ans1+k,新终点同理。

然后跑一边费用流。大功告成!

看代码:

#include<bits/stdc++.h>
using namespace std;
#define inf 1e9
const int maxn=1e5+10;
int n,m,k;
int beg1[maxn],nex1[maxn],to1[maxn],w1[maxn],e1;
int beg2[maxn],nex2[maxn],to2[maxn],w2[maxn],cost[maxn],e2;
inline void add1(int x,int y,int z){
    nex1[e1]=beg1[x];beg1[x]=e1;
    to1[e1]=y;w1[e1]=z;e1++;
}
inline void add2(int x,int y,int z,int c){
    nex2[e2]=beg2[x];beg2[x]=e2;
    to2[e2]=y;w2[e2]=z;cost[e2]=c;e2++;
}
inline int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*f;
}
int dep[maxn],dis[maxn],vis[maxn];
int flow[maxn],las[maxn],ed[maxn];
queue<int>q;
inline int bfs(){
    memset(dep,0x3f,sizeof(dep));
    while(!q.empty())q.pop();
    dep[1]=1;q.push(1);
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=beg1[x];~i;i=nex1[i]){
            int t=to1[i];
            if(w1[i]&&dep[t]>maxn){
                dep[t]=dep[x]+1;
                q.push(t);
            }
        }
    }
    return dep[n]<=maxn;
}
inline int dfs(int x,int lim){
    if(x==n||!lim)return lim;
    int ans=0;
    for(int i=beg1[x];~i;i=nex1[i]){
        int t=to1[i];
        if(w1[i]&&dep[t]==dep[x]+1){
            int f=dfs(t,min(w1[i],lim));
            ans+=f;lim-=f;
            w1[i]-=f;w1[i^1]+=f;
        }
    }
    return ans;
}
inline int spfa(){
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    while(!q.empty())q.pop();
    flow[0]=inf,dis[0]=0,vis[0]=1;
    las[0]=inf,las[n+1]=-1,q.push(0);
    while(!q.empty()){
        int x=q.front();
        q.pop();vis[x]=0;
        for(int i=beg2[x];~i;i=nex2[i]){
            int t=to2[i];
            if(w2[i]&&dis[t]>dis[x]+cost[i]){
                dis[t]=dis[x]+cost[i];
                las[t]=x;ed[t]=i;
                flow[t]=min(flow[x],w2[i]);
                if(!vis[t]){
                    vis[t]=1;
                    q.push(t);
                }
            }
        }
    }
    return las[n+1]!=-1;
}
inline int calc(){
    int tmp=n+1;
    while(tmp){
        w2[ed[tmp]]-=flow[n+1];
        w2[ed[tmp]^1]+=flow[n+1];
        tmp=las[tmp];
    }
    return flow[n+1]*dis[n+1];
}
int main(){
    memset(beg1,-1,sizeof(beg1));
    memset(beg2,-1,sizeof(beg2));
    n=read(),m=read(),k=read();
    int x,y,z,c;
    for(int i=1;i<=m;i++){
        x=read(),y=read(),z=read(),c=read();
        add1(x,y,z),add1(y,x,0);
        add2(x,y,z,0),add2(y,x,0,0);
        add2(x,y,inf,c),add2(y,x,0,-c);
    }
    int maxflow=0,mincost=0;
    while(bfs())maxflow+=dfs(1,inf);
    add2(0,1,maxflow+k,0),add2(1,0,0,0);
    add2(n,n+1,maxflow+k,0),add2(n+1,n,0,0);
    while(spfa())mincost+=calc();
    printf("%d %d
",maxflow,mincost);
    return 0;
} 

罕见的网络流写上100+行。

原文地址:https://www.cnblogs.com/syzf2222/p/12639572.html