有关网络流的一些板子题

最大流 (洛谷 3376)

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }

const int maxn=520010;
const ll INF=2005020600;
int head[maxn];
int n,m,s,t,u,v;
int now[maxn],cnt=1;
ll dis[maxn],ans,w;

struct edge {
    int v,next;
    ll val;
    // edge (int v=0,int next=0,ll val=0): v(v), next(next), val(val) {}
}e[maxn];

inline void add (int u,int v,ll w) {
    e[++cnt].v=v;
    e[cnt].val=w;            //加正反两条边
    e[cnt].next=head[u];
    head[u]=cnt;

    e[++cnt].v=u;
    e[cnt].val=0;
    e[cnt].next=head[v];
    head[v]=cnt;
}

inline int bfs () {          //求分层图
    rep (i,1,n) dis[i]=INF;
    queue <int > q;
    dis[s]=0;
    now[s]=head[s];
    q.push (s);
    while (q.size ()) {
        int x=q.front (); q.pop();          
        for (int i=head[x];i;i=e[i].next) {       
            int v=e[i].v;
            if (e[i].val>0&&dis[v]==INF) {     //直接遍历邻边判断即可
                q.push (v);
                now[v]=head[v];
                dis[v]=dis[x]+1;
                if (v==t) return 1;
            }
        }

    }
    return 0;
}

inline ll dfs (int x,ll sum) {
    if (x==t) return sum;
    ll k,res=0;
    for (int i=now[x];i&&sum;i=e[i].next) {
        now[x]=i;
        int v=e[i].v;
        if (e[i].val>0&&(dis[v]==dis[x]+1)) {
            k=dfs (v,min (sum,e[i].val));      //剩余流量最小值
            if (k==0) dis[v]=INF;
            e[i].val-=k;            //更新残量网络
            e[i^1].val+=k;
            res+=k;         //更新当前答案和剩余流量
            sum-=k;
        }
    }
    return res;
}

int main () {
//    freopen ("data.in","r",stdin);
   scanf ("%d%d%d%d",&n,&m,&s,&t);
   rep (i,1,m) {
       scanf ("%d%d%lld",&u,&v,&w);
       add (u,v,w);
   } 
   while (bfs ()) {         //如果还存在合法分层图,累加答案
       ans+=dfs (s,INF);
   }
   printf ("%lld",ans);
//    fclose (stdin);
   return 0;
}

最小费用最大流

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f = -1;
        ch=getchar();
    } 
    while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }   return x*f;
}

const int maxn=5e4+10;
const int M=5e3+10;
int n,m,s,t,cnt=1;
int head[maxn],dis[maxn],pre[maxn],incf[maxn];
int maxflow,mincost;
bool vis[maxn];
struct edge {
    int v,next,flow,cost;
}e[M<<1];

inline void add (int u,int v,int flow,int cost) {
    e[++cnt]= (edge) {v,head[u],flow,cost};
    head[u]=cnt;
}

inline void add_edge (int u,int v,int flow,int cost) {
   add (u,v,flow,cost);
   add (v,u,0,-cost);
}

inline bool spfa () {
    queue <int > q;
    MT (dis,0x3f);
    MT (vis,0);
    q.push (s);
    dis[s]=0;
    vis[s]=1;
    incf[s]=1<<30;
    while (q.size ()) {
        int x=q.front (); q.pop ();
        vis[x]=0;
        for (int i=head[x];i;i=e[i].next) {
           if (!e[i].flow) continue;
           int v=e[i].v;
           if (dis[v]>dis[x]+e[i].cost) {
               dis[v]=dis[x]+e[i].cost;
               incf[v]=min (incf[x],e[i].flow);
               pre[v]=i;
               if (!vis[v]) vis[v]=1,q.push (v);
           }
        }
    }
    if (dis[t]==inf) return 0;
    return dis[t];
}

inline void mcmf () {
    while (spfa ()) {
       int x=t;
       maxflow+=incf[t];
       mincost+=dis[t]*incf[t];
       int i;
       while (x!=s) {
           i=pre[x];
           e[i].flow-=incf[t];
           e[i^1].flow+=incf[t];
           x=e[i^1].v;
       }
    }
}

int main () {
    scanf ("%d%d%d%d",&n,&m,&s,&t);
    rep (i,1,m) {
        int u,v,w,c;
        scanf ("%d%d%d%d",&u,&v,&w,&c);
        add_edge (u,v,w,c);
    }
    mcmf ();
    printf ("%d %d\n",maxflow,mincost);
    return 0;
}

卡Dinic毒瘤题专用ISAP (hdu 4280)

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}

const int maxn=1e5+10;
struct Edge {
    int u,v;
    ll cap;
    Edge (int u=0,int v=0,ll cap=0) :u (u),v (v),cap (cap) {}
}edge[maxn*4];
vector <int> G[maxn];
int dis[maxn],cur[maxn],num[maxn],pre[maxn],cnt,s,t;

void add_edge (int u,int v,int w) {
    edge[cnt]=Edge (u,v,w);
    G[u].pb (cnt++);
    edge[cnt]=Edge (v,u,w);
    G[v].pb (cnt++);
}

int bfs () {
    MT (dis,-1);
    queue <int> q; q.push (t);
    dis[t]=0;
    while (q.size ()) {
        int u=q.front (); q.pop ();
        for (auto it:G[u]) {
            Edge &e=edge[it];
            if (dis[e.v]==-1) {
                dis[e.v]=dis[u]+1;
                q.push (e.v);
            }
        }
    }
    return ~dis[s];
}

int dfs () {
    int u=t; int flow=inf;
    while (u!=s) {
        Edge &e=edge[pre[u]];
        if (e.cap<flow) flow=e.cap;
        u=e.u;
    } u=t;
    while (u!=s) {
        Edge &e1=edge[pre[u]];
        Edge &e2=edge[pre[u]^1];
        e1.cap-=flow; e2.cap+=flow;
        u=e1.u;
    }
    return flow;
}

int isap (int n) {
    if (!bfs ()) return 0;
    MT (num,0);
    MT (cur,0);
    rep (i,1,n) if (~dis[i]) num[dis[i]]++;
    int ans=0,u=s;
    while (dis[s]<n) {
        if (u==t) ans+=dfs (),u=s;
        int flag=0;
        for (auto it:G[u]) {
            Edge &e=edge[it];
            if (dis[e.v]+1==dis[u]&&e.cap>0) {
                pre[e.v]=it; cur[u]=it;
                flag=1; u=e.v;
                break;
            }
        }
        if (!flag) {
            if (--num[dis[u]]==0) break;
            int md=n;
            for (auto it:G[u]) {
                Edge &e=edge[it];
                if (e.cap>0&&dis[e.v]<md) {
                    md=dis[e.v]; cur[u]=it;
                }
            }
            num[dis[u]=md+1]++;
            if (u!=s) u=edge[pre[u]].u;
        }
    }
    return ans;
}

int main () {
    int tt;
    scanf ("%d",&tt);
    while (tt--) {
        int n,m;
        scanf ("%d%d",&n,&m);
        cnt=0; rep (i,0,n) G[i].clear ();
        int u,v,c,west=inf,east=-inf;
        rep (i,1,n) {
            scanf ("%d%d",&u,&v);
            if (west>u) west=u,s=i;
            if (east<u) east=u,t=i;
        }
        rep (i,1,m) {
            scanf ("%d%d%d",&u,&v,&c);
            add_edge (u,v,c);
        }
        printf ("%d\n",isap (n));
    }


    return 0;
}
posted @ 2020-08-20 22:36  Luglucky  阅读(157)  评论(0编辑  收藏  举报