2017江苏省省赛 Roads（全局最小割）

Roads

时间限制: 2 Sec  内存限制: 256 MB
提交: 6  解决: 2
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题目描述

In ICPCCamp, there are n towns conveniently labeled with 1,2,…,n and m bidirectional roads planned to be built. The i-th road will be built between cities ai and bi with cost ci. The builders in ICPCCamp will build the (n−1) roads with the least total cost to connect any of two cities directly or indirectly.

Bobo, the mayor of ICPCCamp is going to remove some of the roads from the construction plan. He would like to know the minimum number roads to be removed to strictly increases the total cost.

Note that the total cost is considered as +∞ if no valid (n−1) roads exist after removing. It is also counted as "total cost strictly increases".

输入

The input contains zero or more test cases and is terminated by end-of-file. For each test case:
The first line contains two integers n and m. The i-th of the following m lines contains ai,bi,ci.
2≤n≤50,n−1≤m≤n2,1≤ai,bi≤n,1≤ci≤109
Any two cities will be connected if all m roads are built.
The sum of n does not exceed 103.

输出

For each case, output an integer which denotes the result.

样例输入

3 3
1 2 1
1 3 2
2 3 3
3 4
1 2 1
1 2 1
1 3 2
1 3 3
3 4
1 2 1
1 2 1
1 3 2
1 3 2
4 6
1 2 1
1 3 1
1 4 1
2 3 1
2 4 1
3 4 1

样例输出

1
1
2
3

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e2+50;;
const int M = 17;
const int mod = 19260817;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
ll qpow(int x,int qq){ll f=1,p=x;while(qq){if(qq&1)f=f*p%mod;p=1LL*p*p%mod;qq>>=1;}}
int n,m;
int fa[N],wage[N],edg[N][N];
bool vis[N],in_set[N];
struct man{
     int u,v,w;
     bool operator <(const man &s)const {
         return w<s.w;
     }
}a[N*N];
int findFa(int x){
    return fa[x]==x?x:fa[x]=findFa(fa[x]);
}
void unionFa(int x,int y){
    if(findFa(x)==findFa(y))return ;
    fa[findFa(x)]=findFa(y);
}
int search(int &s,int &t){
    met(vis,false);
    met(wage,0);
    int u,mincut,mx;
    for(int i=1;i<=n;i++){
        u=mx=-1;
        for(int j=1;j<=n;j++){
            if(!in_set[j]&&!vis[j]&&wage[j]>mx)mx=wage[u=j];
        }
        if(u==-1)return mincut;
        vis[u]=true;
        s=t,t=u;
        mincut=mx;
        for(int j=1;j<=n;j++){
            if(!in_set[j]&&!vis[j])wage[j]+=edg[u][j];
        }
    }
    return mincut;
}
int stoer_wagner(){
    int  mincut=inf,s,t,ret;
    for(int i=1;i<n;i++){
        ret=search(s,t);
        in_set[t]=true;
        if(ret!=0&&ret<mincut)mincut=ret;//加ret!=0是为了求所有联通块的最小割
        for(int j=1;j<=n;j++)if(!in_set[j])edg[s][j]=(edg[j][s]+=edg[j][t]);
    }
    return mincut;
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<=n;i++)fa[i]=i;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].w);
        }
        sort(a,a+m);
        int ans=inf,cnt=0;
        for(int i=0;i<m;){
            int now=i,w=a[i].w;
            met(edg,0);met(in_set,false);
            for(;a[i].w==w&&i<m;i++){
                if(findFa(a[i].u)!=findFa(a[i].v)){
                    edg[findFa(a[i].u)][findFa(a[i].v)]++;
                    edg[findFa(a[i].v)][findFa(a[i].u)]++;
                }
            }
            ans=min(ans,stoer_wagner());
            for(int i=now;a[i].w==w&&i<m;i++){
                if(findFa(a[i].u)!=findFa(a[i].v)){
                    unionFa(a[i].u,a[i].v);
                    cnt++;
                }
            }
            if(cnt>=n-1)break;
        }
        printf("%d\n",ans);
    }
    return 0;
}