zoukankan      html  css  js  c++  java
  • 最大流模板

    ISAP+BFS初始化+栈优化

    跑的飞快,但比较长

    点数最大大概(10^5),边数最大大概(4 imes 10^5)

    #include <bits/stdc++.h>
    using namespace std;
    const int inf = 0x3f3f3f3f;
    const int N = 60050;
    struct node {
        int v, cap, flow, nxt;
        node () {}
        node (int v, int cap, int flow, int nxt): v(v), cap(cap), flow(flow), nxt(nxt) {}
    } edge[N * 10];
    int head[N], tot;
    void init() {
        memset(head, -1, sizeof(head));
        tot = 0;
    }
    int dep[N];
    void adde(int u, int v, int w) {
        edge[tot] = node(v, w, 0, head[u]);
        head[u] = tot++;
        edge[tot] = node(u, 0, 0, head[v]);
        head[v] = tot++;
    }
    int gap[N], cur[N];
    void bfs(int s, int t) {
        memset(dep, -1, sizeof(dep));
        memset(gap, 0, sizeof(gap));
        gap[0] = 1;
        queue<int> q;
        dep[t] = 0;
        q.push(t);
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (int i = head[u]; ~i; i = edge[i].nxt) {
                int v = edge[i].v;
                if (dep[v] != -1) continue;
                q.push(v);
                dep[v] = dep[u] + 1;
                gap[dep[v]]++;
            }
        }
    }
    int S[N];
    int sap(int s, int t, int n) {
        bfs(s, t);
        memcpy(cur, head, sizeof(head));
        int top = 0;
        int u = s;
        int ans = 0;
        while (dep[s] < n) {
            if (u == t) {
                int minn = inf;
                int inser;
                for (int i = 0; i < top; i++) {
                    if (minn > edge[S[i]].cap - edge[S[i]].flow) {
                        minn = edge[S[i]].cap - edge[S[i]].flow;
                        inser = i;
                    }
                }
                for (int i = 0; i < top; i++) {
                    edge[S[i]].flow += minn;
                    edge[S[i] ^ 1].flow -= minn;
                }
                ans += minn;
                top = inser;
                u = edge[S[top] ^ 1].v;
                continue;
            }
            bool flag = false;
            int v;
            for (int i = cur[u]; ~i; i = edge[i].nxt) {
                v = edge[i].v;
                if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]) {
                    flag = true;
                    cur[u] = i;
                    break;
                }
            }
            if (flag) {
                S[top++] = cur[u];
                u = v;
                continue;
            }
            int minn = n;
            for (int i = head[u]; ~i; i = edge[i].nxt) {
                if (edge[i].cap - edge[i].flow && dep[edge[i].v] < minn) {
                    minn = dep[edge[i].v];
                    cur[u] = i;
                }
            }
            gap[dep[u]]--;
            if (!gap[dep[u]]) return ans;
            dep[u] = minn + 1;
            gap[dep[u]]++;
            if (u != s) u = edge[S[--top] ^ 1].v;
        }
        return ans;
    }
    

    普通ISAP 邻接表形式

    跑的也挺快的,大概所需时间是dinic的2/3,无递归,可以完美替代dinic

    点数最大大概(10^5),边数最大大概(4 imes 10^5)

    #include <bits/stdc++.h>
    using namespace std;
    const int inf = 0x3f3f3f3f;
    const int N = 20050;
    struct node {
        int v, cap, nxt;
        node () {}
        node (int v, int cap, int nxt): v(v), cap(cap), nxt(nxt) {}
    } edge[N * 100];
    int head[N], tot;
    void init() {
        memset(head, -1, sizeof(head));
        tot = 0;
    }
    int dep[N];
    void adde(int u, int v, int w) {
        edge[tot] = node(v, w, head[u]);
        head[u] = tot++;
        edge[tot] = node(u, 0, head[v]);
        head[v] = tot++;
    }
    int cur[N];
    int gap[N];
    int pre[N];
    int sap(int s, int t, int n) {
        memset(dep, 0, sizeof(dep));
        memset(gap, 0, sizeof(gap));
        memcpy(cur, head, sizeof(head));
        memset(pre, 0, sizeof(pre));
        gap[0] = n;
        int u = s;
        pre[u] = -1;
        int ans = 0;
        while (dep[s] < n) {
            if (u == t) {
                int minn = inf;
                for (int i = pre[u]; ~i; i = pre[edge[i ^ 1].v]) {
                    if (minn > edge[i].cap) minn = edge[i].cap;
                }
                for (int i = pre[u]; ~i; i = pre[edge[i ^ 1].v]) {
                    edge[i].cap -= minn;
                    edge[i ^ 1].cap += minn;
                }
                u = s;
                ans += minn;
                continue;
            }
            bool flag = false;
            int v;
            for (int i = cur[u]; ~i; i = edge[i].nxt) {
                v = edge[i].v;
                if (edge[i].cap && dep[v] + 1 == dep[u]) {
                    flag = true;
                    cur[u] = pre[v] = i;
                    break;
                }
            }
            if (flag) {
                u = v;
                continue;
            }
            int minn = n;
            for (int i = head[u]; ~i; i = edge[i].nxt) {
                if (edge[i].cap && dep[edge[i].v] < minn) {
                    minn = dep[edge[i].v];
                    cur[u] = i;
                }
            }
            gap[dep[u]]--;
            if (!gap[dep[u]]) return ans;
            dep[u] = minn + 1;
            gap[dep[u]]++;
            if (u != s) u = edge[pre[u] ^ 1].v;
        }
        return ans;
    }
    

    还有一种递归写法,挺慢的,和dinic差不多,还不如dinic好写,没啥用。

    Dinic

    大部分时间够用,但普通ISAP只比dinic长10行,代码量相差不大。

    #include <bits/stdc++.h>
    using namespace std;
    const int inf = 0x3f3f3f3f;
    const int N = 60050;
    struct node {
        int u, v, cap, nxt;
        node () {}
        node (int u, int v, int cap, int nxt): u(v), v(v), cap(cap), nxt(nxt) {}
    } edge[N * 10];
    int head[N], tot;
    void init() {
        memset(head, -1, sizeof(head));
        tot = 0;
    }
    int dep[N];
    void adde(int u, int v, int w) {
        edge[tot] = node(u, v, w, head[u]);
        head[u] = tot++;
        edge[tot] = node(v, u, 0, head[v]);
        head[v] = tot++;
    }
    bool bfs(int s, int t) {
        queue<int> q;
        memset(dep, -1, sizeof(dep));
        dep[s] = 0;
        q.push(s);
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (int i = head[u]; ~i; i = edge[i].nxt) {
                int v = edge[i].v;
                if (edge[i].cap > 0 && dep[v] == -1) {
                    dep[v] = dep[u] + 1;
                    q.push(v);
                }
            }
        }
        return dep[t] > 0;
    }
    int dfs(int u, int t, int f) {//f是到这点的残余流量,used是这点已经流出的流量
        if (u == t) return f;
        int w, used = 0;
        for (int i = head[u]; ~i; i = edge[i].nxt) {
            int v = edge[i].v;
            int cap = edge[i].cap;
            if (cap && dep[v] == dep[u] + 1) {
                w = f - used;
                w = dfs(v, t, min(w, cap));
                edge[i].cap -= w;
                edge[i ^ 1].cap += w;
                used += w;
                if (used == f) return f;
            }
        }
        if (!used) dep[u] = -1;
        return used;
    }
    int dinic(int s, int t) {
        int maxflow = 0;
        while (bfs(s, t)) {
           maxflow += dfs(s, t, inf);
        }
        return maxflow;
    }
    

    最高标号预流推进(HLPP)

    我觉得可能用不到把,复杂度是(n^2 sqrt{m}),但是常数比较大,和isap跑的差不多快

    给个别人的板子,自己没写

    #include <bits/stdc++.h>
    using namespace std;
    inline int Read(){
        int x=0;
        char c=getchar();
        while(c>'9'||c<'0')c=getchar();
        while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
        return x;
    }
    const int inf=1<<30;
    int top=1,head[10100];
    int n,m,s,t;
    int e[10100],h[10100],cnth[20100];//每个点对应的余流,高度;每个高度有多少个点 
    struct cmp{
        inline bool operator () (int a,int b) const{
            return h[a]<h[b];
        }
    };
    struct Node{
        int v;
        int val;
        int next;
    }node[400100];
    inline void addedge(int u,int v,int val){
        node[++top].v=v;
        node[top].val=val;
        node[top].next=head[u];
        head[u]=top;
    }
    inline void add(int u,int v,int val){
        addedge(u,v,val);
        addedge(v,u,0);
    }
    int inque[11000];
    void bfs(){
        memset(h,0x3f,sizeof(h));
        h[t]=0;
        queue<int>qu;
        qu.push(t);
        while(!qu.empty()){
            int u=qu.front();
            qu.pop();
            inque[u]=0;
            for(int i=head[u];i;i=node[i].next){
                int d=node[i].v;
                if(node[i^1].val&&h[d]>h[u]+1){//反向跑 
                    h[d]=h[u]+1;
                    if(inque[d]==0){
                        qu.push(d);
                        inque[d]=1;
                    }
                }
            }
        }
        return;
    }
    priority_queue<int,vector<int>,cmp>q;
    inline void push_(int u){
        for(int i=head[u];i;i=node[i].next){
            int d=node[i].v;
            if(node[i].val&&h[d]+1==h[u]){//可以推流 
                int mi=min(node[i].val,e[u]);
                node[i].val-=mi;
                node[i^1].val+=mi;
                e[u]-=mi;
                e[d]+=mi;
                if(inque[d]==0&&d!=t&&d!=s){
                    q.push(d);
                    inque[d]=1;
                }
                if(e[u]==0)break;//已经推完了 
            }
        }
    }//推流 
    inline void relabel(int u){
        h[u]=inf;
        for(int i=head[u];i;i=node[i].next){
            int d=node[i].v;
            if(node[i].val&&h[d]+1<h[u]){
                h[u]=h[d]+1;
            }
        }
    }//把u的高度更改为与u相邻的最低的点的高度加1 
    int hlpp(){
        register int i;
        bfs();
        if(h[s]==0x3f3f3f3f)return 0;//s与t不连通 
        h[s]=n;
        for(i=1;i<=n;i++)if(h[i]<0x3f3f3f3f)cnth[h[i]]++;//统计各个高度的点数,注意不要让下标越界
        for(i=head[s];i;i=node[i].next){
            int d=node[i].v;    
            int mi=node[i].val;
            if(mi){
                e[s]-=mi;
                e[d]+=mi;
                node[i].val-=mi;
                node[i^1].val+=mi;
                if(d!=t&&inque[d]==0&&d!=s){
                q.push(d);
                inque[d]=1; 
                }
            }
        }//从s向周围点推流
        while(!q.empty()){
            int u=q.top();
            inque[u]=0;
            q.pop();
            push_(u);
            if(e[u]){//还有余流 
                cnth[h[u]]--;
                if(cnth[h[u]]==0){
                    for(int i=1;i<=n;i++){
                        if(i!=s&&i!=t&&h[i]>h[u]&&h[i]<n+1){
                            h[i]=n+1;//标记无法到达 
                        }
                    }
                }//gap优化 
                relabel(u);
                cnth[h[u]]++;
                q.push(u);
                inque[u]=1;
            }
        }
        return e[t];
    } 
    int main(){
        n=Read(),m=Read(),s=Read(),t=Read();
        register int i;
        int u,v,val;
        for(i=1;i<=m;i++)u=Read(),v=Read(),val=Read(),add(u,v,val);
        printf("%d",hlpp());
        return 0;
    }
    
  • 相关阅读:
    刚刚学习Silverlight
    给文本框添加水印效果
    .net 下实现下载
    UpdatePanel中弹出对话框
    用VS.NET开发在Linux Apache Tomcat上运行的应用
    玩儿条形码之条码生成
    关于ContextSwitchDeadlock
    第一个Grasshoper应用
    WebService实现Ajax
    使用decorator的线程同步
  • 原文地址:https://www.cnblogs.com/artoriax/p/12203388.html
Copyright © 2011-2022 走看看