【CQOI2016】不同的最小割 【分治+最大流最小割/最小割树】

LibreOJ - 2042 题目链接

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  这道题有个可能TLE的坑点，就是我们按照正常的网络流的建边，就是u->v(flow), v->u(0)、u->v(0), v->u(flow)。这样的无向图的建边方式，在这里会TLE的，这里算是卡了这个2倍的常数，可能会过不了最后一组样例。

  剩下的，我们可以利用分治法加上最大流，以复杂度为的复杂度来解决这个问题。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 855, maxM = 9e3 + 7;
int N, M, head[maxN], cnt;
struct Eddge
{
    int nex, to, flow;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), flow(c) {}
}edge[maxM << 2];
inline void addEddge(int u, int v, int flow)
{
    edge[cnt] = Eddge(head[u], v, flow); head[u] = cnt++;
    edge[cnt] = Eddge(head[v], u, flow); head[v] = cnt++;
}
inline void _add(int u, int v, int flow) { addEddge(u, v, flow); /*addEddge(v, u, flow);*/ }
struct Max_Flow
{
    int deep[maxN], cur[maxN], S, T;
    queue<int> Q;
    inline bool bfs()
    {
        while(!Q.empty()) Q.pop();
        for(int i=1; i<=N; i++) deep[i] = 0;
        deep[S] = 1;
        Q.push(S);
        int u;
        while(!Q.empty())
        {
            u = Q.front(); Q.pop();
            for(int i=head[u], v, f; ~i; i=edge[i].nex)
            {
                v = edge[i].to; f = edge[i].flow;
                if(f && !deep[v])
                {
                    deep[v] = deep[u] + 1;
                    Q.push(v);
                }
            }
        }
        return deep[T];
    }
    int dfs(int u, int dist)
    {
        if(u == T) return dist;
        for(int &i=cur[u], v, f, di; ~i; i=edge[i].nex)
        {
            v = edge[i].to; f = edge[i].flow;
            if(f && deep[v] == deep[u] + 1)
            {
                di = dfs(v, min(dist, f));
                if(di)
                {
                    edge[i].flow -= di; edge[i ^ 1].flow += di;
                    return di;
                }
            }
        }
        return 0;
    }
    inline void Init()
    {
        for(int i=0; i<cnt; i+=2)
        {
            edge[i].flow = edge[i ^ 1].flow = (edge[i].flow + edge[i ^ 1].flow) >> 1;
        }
    }
    inline int Dinic(int st, int ed)
    {
        int ans = 0, tmp;
        S = st; T = ed;
        Init();
        while(bfs())
        {
            for(int i=1; i<=N; i++) cur[i] = head[i];
            while((tmp = dfs(S, INF))) ans += tmp;
        }
        return ans;
    }
}flow;
int Lsan[maxN], _UP = 0, p[maxN], cop[maxN];
void cdq(int l, int r)
{
    if(l == r) return;
    Lsan[++_UP] = flow.Dinic(p[l], p[l + 1]);
    int ql = l - 1, qr = r + 1;
    for(int i=l; i<=r; i++)
    {
        if(flow.deep[p[i]]) cop[++ql] = p[i];
        else cop[--qr] = p[i];
    }
    for(int i=l; i<=r; i++) p[i] = cop[i];
    cdq(l, ql); cdq(qr, r);
}
inline void init()
{
    cnt = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &M);
    init();
    for(int i=1, u, v, w; i<=M; i++)
    {
        scanf("%d%d%d", &u, &v, &w);
        _add(u, v, w);
    }
    for(int i=1; i<=N; i++) p[i] = i;
    cdq(1, N);
    sort(Lsan + 1, Lsan + _UP + 1);
    printf("%d\n", (int)(unique(Lsan + 1, Lsan + _UP + 1) - Lsan - 1));
    return 0;
}