【hihocoder 1378】网络流二·最大流最小割定理

【Link】:http://hihocoder.com/problemset/problem/1378

【Description】

【Solution】



在求完最小割(最大流)之后;

可以在剩余网络中再从1号点做一次bfs;

往flow[][]为正的边走;

能走到的点就是S集合了;



【NumberOf WA】

【Reviw】

【Code】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x+1)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 500;
const int INF = 0x3f3f3f3f;

int n,m,flow[N+10][N+10],pre[N+10],ans1,ans2;
queue <int> dl;
vector <int> v;

int main(){
    //Open();
    //Close();
    ri(n),ri(m);
    rep1(i,1,m){
        int x,y,z;
        ri(x),ri(y),ri(z);
        flow[x][y]+=z;
    }

    ans1 = 0;
    while (1){
        ms(pre,255);
        dl.push(1);
        pre[1] = 0;
        while (!dl.empty()){
            int x = dl.front();
            dl.pop();
            for (int i = 1;i <= n;i++)
                if (pre[i]==-1 && flow[x][i]){
                    pre[i] = x;
                    dl.push(i);
                }
        }
        if (pre[n]==-1) break;
        int mi = INF;
        int now = n;
        while (now != 1){
            mi = min(mi,flow[pre[now]][now]);
            now = pre[now];
        }
        now = n;
        while (now != 1){
            flow[pre[now]][now] -= mi;
            now = pre[now];
        }
        ans1 += mi;
    }

    ms(pre,0);
    dl.push(1);pre[1] = 1;
    while (!dl.empty()){
        int x = dl.front();dl.pop();
        rep1(i,1,n)
            if (pre[i]==0 && flow[x][i]){
                pre[i] = 1;
                dl.push(i);
            }
    }

    ans2 = 0;

    rep1(i,1,n)
        if (pre[i]){
            ans2++;
            v.pb(i);
        }
    oi(ans1);oc;oi(ans2);puts("");
    int len = v.size();
    rep1(i,0,len-1){
        oi(v[i]);
        if (i==len-1)
            puts("");
        else
            oc;
    }
    return 0;
}