HDU 3879 Base Station 最大权闭合图

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题目链接：点击打开链接

题意：

给定n个带权点m条无向带权边

选一个子图，则这个子图的权值为 边权和-点权和

求一个最大的权值

把边也当成点。然后是最大权闭合图

dinic:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
//点标 从0开始 F.Init(n) n=最大点标+10
const int N = 200010;
const int M = 500010;
const int INF = ~0u >> 2;
template<class T>
struct Max_Flow {
    int n;
    int Q[N], sign;
    int head[N], level[N], cur[N], pre[N];
    int nxt[M], pnt[M], E;
    T cap[M];
    void Init(int n) {
        this->n = n;
        E = 0;
        std::fill(head, head + n, -1);
    }
    //有向rw 就= 0
    void add(int from, int to, T c, T rw) {
        pnt[E] = to;
        cap[E] = c;
        nxt[E] = head[from];
        head[from] = E++;

        pnt[E] = from;
        cap[E] = rw;
        nxt[E] = head[to];
        head[to] = E++;
    }
    bool Bfs(int s, int t) {
        sign = t;
        std::fill(level, level + n, -1);
        int *front = Q, *tail = Q;
        *tail++ = t; level[t] = 0;
        while(front < tail && level[s] == -1) {
            int u = *front++;
            for(int e = head[u]; e != -1; e = nxt[e]) {
                if(cap[e ^ 1] > 0 && level[pnt[e]] < 0) {
                    level[pnt[e]] = level[u] + 1;
                    *tail ++ = pnt[e];
                }
            }
        }
        return level[s] != -1;
    }
    void Push(int t, T &flow) {
        T mi = INF;
        int p = pre[t];
        for(int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {
            mi = std::min(mi, cap[p]);
        }
        for(int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {
            cap[p] -= mi;
            if(!cap[p]) {
                sign = pnt[p ^ 1];
            }
            cap[p ^ 1] += mi;
        }
        flow += mi;
    }
    void Dfs(int u, int t, T &flow) {
        if(u == t) {
            Push(t, flow);
            return ;
        }
        for(int &e = cur[u]; e != -1; e = nxt[e]) {
            if(cap[e] > 0 && level[u] - 1 == level[pnt[e]]) {
                pre[pnt[e]] = e;
                Dfs(pnt[e], t, flow);
                if(level[sign] > level[u]) {
                    return ;
                }
                sign = t;
            }
        }
    }
    T Dinic(int s, int t) {
        pre[s] = -1;
        T flow = 0;
        while(Bfs(s, t)) {
            std::copy(head, head + n, cur);
            Dfs(s, t, flow);
        }
        return flow;
    }
};
Max_Flow <int>F;
int n, m;
int work(){
    F.Init(n+m+10);
    int from = 0, to = n + m +1, A;
    for(int i = 1; i <= n; i++){
        scanf("%d", &A);
        F.add(i, to, A, 0);
    }
    int u, v, d, all = 0;
    for(int i = 1; i <= m; i++){
        scanf("%d %d %d", &u, &v, &d);
        all += d;
        F.add(from, n+i, d, 0);
        F.add(n+i, u, INF, 0);
        F.add(n+i, v, INF, 0);
    }
    return all - F.Dinic(from, to);
}
int main(){
    while(cin>>n>>m)
        cout<<work()<<endl;
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define ll int
const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void add(int u,int v,int w,int rw = 0)
{
    edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0;
    edge[tol].next = head[u]; head[u] = tol++;
    edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0;
    edge[tol].next = head[v]; head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dep[v] != -1)continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end,int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            int inser;
            for(int i = 0;i < top;i++)
                if(Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            for(int i = 0;i < top;i++)
            {
                edge[S[i]].flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            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 Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)u = edge[S[--top]^1].to;
    }
    return ans;
}
void init(){ tol = 0; memset(head,-1,sizeof(head)); }

int n, m;
ll work(){
    init();
    int from = 0, to = n + m +1, A;
    for(int i = 1; i <= n; i++){
        scanf("%d", &A);
        add(i, to, A);
    }
    int u, v; ll d;
    ll all = 0;
    for(int i = 1; i <= m; i++){
        scanf("%d %d %d", &u, &v, &d);
        all += d;
        add(from, n+i, d);
        add(n+i, u, INF);
        add(n+i, v, INF);
    }
    return all - sap(from, to,to+1);
}
int main(){
    while(cin>>n>>m)
        cout<<work()<<endl;
    return 0;
}

HDU 3879 Base Station 最大权闭合图

标签：blog   http   io   ar   os   sp   for   on   2014   

原文地址：http://blog.csdn.net/qq574857122/article/details/41093153