我们有某些结论,本质不同的最小割一共有n-1个。
在这颗最小割树上,我们有两种点集,一种是源点点集,一种是汇点点集
我们做一次dinic后被增广到的地方就属于源点点集,否则属于汇点点集。这两个点集之间我们任意选的s和t之间的连边就是最小割的大小
然后我们分治递归两个子树来构建这颗最小割树
性质还有任意两个点之间的路径的最小权值就是这两点的最小割
然后这就成为了分治最小割的裸题了
/* ***********************************************
Author :BPM136
Created Time :2016/4/20 14:54:49
File Name :A.cpp
************************************************ */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#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){
queueq;
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;
}