BZOJ 1834 网络扩容(最大流+费用流)

对于第一问，直接求最大流。

对于第二问，建源点s和汇点t，s连1容量为INF,费用为0的边，n连t容量为最大流+k，费用为0的边。这样就把最大流限制为最多增加k了。

限制需要求扩充的最小费用，原图的边多连一条容量为INF,费用为增容费用K的边。跑一遍费用流即是答案。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 1000000009
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1005;
//Code begin...

struct Edge{int p, next, w;}edge[N*10];
struct Edge1{int to, next, cap, flow, cost;}edge1[N*20];
struct Node{int u, v, C, K;}node[5005];
int head[N], cnt=2, s, t;
int vis[N], head1[N], tol, pre[N], dis[N];
bool mark[N];
queue<int>Q;

void add_edge(int u, int v, int w){
    edge[cnt].p=v; edge[cnt].next=head[u]; edge[cnt].w=w; head[u]=cnt++;
    edge[cnt].p=u; edge[cnt].next=head[v]; edge[cnt].w=0; head[v]=cnt++;
}
void addedge(int u, int v, int cap, int cost){
    edge1[tol].to=v; edge1[tol].cap=cap; edge1[tol].cost=cost;
    edge1[tol].flow=0; edge1[tol].next=head1[u]; head1[u]=tol++;
    edge1[tol].to=u; edge1[tol].cap=0; edge1[tol].cost=-cost;
    edge1[tol].flow=0; edge1[tol].next=head1[v];  head1[v]=tol++;
}
bool spfa(int s, int t){
    FOR(i,s,t) dis[i]=INF, mark[i]=false, pre[i]=-1;
    dis[s]=0; mark[s]=true; Q.push(s);
    while (!Q.empty()) {
        int u=Q.front(); Q.pop();
        mark[u]=false;
        for (int i=head1[u]; i!=-1; i=edge1[i].next) {
            int v=edge1[i].to;
            if (edge1[i].cap>edge1[i].flow&&dis[v]>dis[u]+edge1[i].cost) {
                dis[v]=dis[u]+edge1[i].cost; pre[v]=i;
                if (!mark[v]) mark[v]=true, Q.push(v);
            }
        }
    }
    if (pre[t]==-1) return false;
    else return true;
}
int minCostMaxflow(int s, int t, int &cost){
    int flow=0;
    cost=0;
    while (spfa(s,t)) {
        int Min=INF;
        for (int i=pre[t]; i!=-1; i=pre[edge1[i^1].to]) {
            Min=min(Min,edge1[i].cap-edge1[i].flow);
        }
        for (int i=pre[t]; i!=-1; i=pre[edge1[i^1].to]) {
            edge1[i].flow+=Min; edge1[i^1].flow-=Min;
            cost+=edge1[i].cost*Min;
        }
        flow+=Min;
    }
    return flow;
}
int bfs(){
    int i, v;
    mem(vis,-1);
    vis[s]=0; Q.push(s);
    while (!Q.empty()) {
        v=Q.front(); Q.pop();
        for (i=head[v]; i; i=edge[i].next) {
            if (edge[i].w>0 && vis[edge[i].p]==-1) {
                vis[edge[i].p]=vis[v] + 1;
                Q.push(edge[i].p);
            }
        }
    }
    return vis[t]!=-1;
}
int dfs(int x, int low){
    int i, a, temp=low;
    if (x==t) return low;
    for (i=head[x]; i; i=edge[i].next) {
        if (edge[i].w>0 && vis[edge[i].p]==vis[x]+1){
            a=dfs(edge[i].p,min(edge[i].w,temp));
            temp-=a; edge[i].w-=a; edge[i^1].w += a;
            if (temp==0) break;
        }
    }
    if (temp==low) vis[x]=-1;
    return low-temp;
}
int main ()
{
    int n, m, k, ans;
    mem(head1,-1);
    scanf("%d%d%d",&n,&m,&k);
    s=0; t=n+1;
    add_edge(s,1,INF); add_edge(n,t,INF);
    FOR(i,1,m) {
        scanf("%d%d%d%d",&node[i].u,&node[i].v,&node[i].C,&node[i].K);
        add_edge(node[i].u,node[i].v,node[i].C);
    }
    int tmp, sum=0;
    while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp;
    addedge(s,1,INF,0); addedge(n,t,sum+k,0);
    FOR(i,1,m) {
        addedge(node[i].u,node[i].v,node[i].C,0);
        addedge(node[i].u,node[i].v,INF,node[i].K);
    }
    minCostMaxflow(s,t,ans);
    printf("%d %d
",sum,ans);
    return 0;
}