CQOI2016 不同的最小割 分治最小割（最小割樹）

我們有某些結論，本質不同的最小割一共有n-1個。

在這顆最小割樹上，我們有兩種點集，一種是源點點集，一種是匯點點集

我們做一次dinic後被增廣到的地方就屬於源點點集，否則屬於匯點點集。這兩個點集之間我們任意選的s和t之間的連邊就是最小割的大小

然後我們分治遞歸兩個子樹來構建這顆最小割樹

性質還有任意兩個點之間的路徑的最小權值就是這兩點的最小割

然後這就成爲了分治最小割的裸題了

/* ***********************************************
Author        :BPM136
Created Time  :2016/4/20 14:54:49
File Name     :A.cpp
************************************************ */

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<bitset>
#include<queue>
#include<ctime>
#include<set>
#include<utility>
#include<vector>
#include<functional>
#include<numeric>
#include<memory>
#include<iterator>
#define LL long long
#define DB double
#define LB long double
#define UL unsigned long
#define ULL unsigned long long
#define get(a,i) a&(1<<(i-1))
#define PAU putchar(0)
#define ENT putchar(32)
#define clr(a,b) memset(a,b,sizeof(a))
#define fo(_i,_a,_b) for(int _i=_a;_i<=_b;_i++)
#define fd(_i,_a,_b) for(int _i=_a;_i>=_b;_i--)
#define efo(_i,_a) for(int _i=last[_a];_i!=0;_i=e[_i].next)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define mkd(x) freopen(#x".in","w",stdout);
#define setlargestack(x) int size=x<<20;char *p=(char*)malloc(size)+size;__asm__("movl %0, %%esp\n" :: "r"(p));
#define end system("pause")
using namespace std;
LL read()
{
         LL f=1,d=0;char s=getchar();
         while (s<48||s>57){if (s==45) f=-1;s=getchar();}
         while (s>=48&&s<=57){d=d*10+s-48;s=getchar();}
         return f*d;
}
LL readln()
{
       LL f=1,d=0;char s=getchar();
       while (s<48||s>57){if (s==45) f=-1;s=getchar();}
       while (s>=48&&s<=57){d=d*10+s-48;s=getchar();}
       while (s!=10) s=getchar();
       return f*d;
}
inline void write(LL x)
{
    if(x==0){putchar(48);return;}if(x<0)putchar(45),x=-x;
    int len=0,buf[20];while(x)buf[len++]=x%10,x/=10;
    for(int i=len-1;i>=0;i--)putchar(buf[i]+48);return;
}
inline void writeln(LL x){write(x);ENT;}
#define N 855
#define M 8505
#define MAXW 100005
const int inf = 0x7fffffff / 2 - 1;
struct edge
{
	int y,next,f;
}e[M*10];
int last[N],ne=1;
int cur[N];
#define efoc(i,x) for(int i=cur[x];i!=0;i=e[i].next)
int n,m;
int Ans[N],Ansnum=0;
int a[N];
int tmp[N];

void add(int x,int y,int f){
	e[++ne].y=y;e[ne].f=f;e[ne].next=last[x];last[x]=ne;
}
void add2(int x,int y,int f){
	add(x,y,f);add(y,x,f);
}

void get_this(){
	for(int k=2;k<=ne;k+=2){
		e[k].f=e[k^1].f=(e[k].f+e[k^1].f)>>1;
	}
}

int high[N];
bool bfs(int s,int t){
	queue<int>q;
	fo(i,0,n)high[i]=-1;
	q.push(s);high[s]=0;
	while(!q.empty()){
		int now=q.front();q.pop();
		efo(i,now){
			if(high[e[i].y]==-1&&e[i].f){
				high[e[i].y]=high[now]+1;
				q.push(e[i].y);
			}
		}
	}
	return high[t]!=-1;
}

int dfs(int x,int f,int T){
	if(x==T||f==0)return f;
	int w,used=0;
	efoc(i,x){
		if(high[e[i].y]==high[x]+1){
			w=f-used;
			w=dfs(e[i].y,min(w,e[i].f),T);
			e[i].f-=w;e[i^1].f+=w;
			used+=w; if(e[i].f) cur[x]=i;
			if(used == f)return used;
		}
	}
	if(!used)high[x]=-1;
	return used;
}

int dinic(int S,int T){
	int ret=0;
	while(bfs(S,T)){
		fo(i,1,n)cur[i]=last[i];
		ret+=dfs(S,inf,T);
	} 
	return ret;
}

bool mark[N];
void dfs_flow(int x){
	mark[x]=1;
	efo(i,x){
		if(mark[e[i].y]==0&&e[i].f){
			dfs_flow(e[i].y);
		}
	}
}

void solve(int l,int r){
	if(l==r)return;
	get_this();
	int ret=dinic(a[l],a[r]);
	Ans[++Ansnum]=ret;
	memset(mark,0,sizeof(mark));
	dfs_flow(a[l]);
	int L=l,R=r;
	fo(i,l,r){
		if(mark[a[i]])tmp[L++]=a[i];
		else tmp[R--]=a[i];
	}
	fo(i,l,r)a[i]=tmp[i];
	solve(l,L-1);solve(R+1,r);
}

int main()
{
	file(A);
	n=read(),m=read();
	fo(i,1,m){
		int x=read(),y=read(),f=read();
		add2(x,y,f);
	}
	fo(i,1,n)a[i]=i;
	solve(1,n);
	sort(Ans+1,Ans+Ansnum+1);
	int ans=1;
	fo(i,2,Ansnum){
		if(Ans[i]!=Ans[i-1])ans++;
	}
	writeln(ans);
	return 0;
}