P2711 小行星 最小割

 

 

如果是一个二维的话很明显是一个二分图最小顶点覆盖

 

三维的话  他说要求消除和最小  所以很明显是一个最小割模型

 

可以将x y z拆点

 

 

并且可以将这张图简化：

 

 

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
const int N=4e5+44;
const int M=4e6+54;
int d[N];
struct edge {
    int to, next, w;
} e[M << 1];
int head[N],cur[N],cnt = 1;
void add(int x, int y, int z) {
    e[++cnt] = (edge){y, head[x], z};
    head[x] = cnt;
    e[++cnt] = (edge){x, head[y], 0};
    head[y] = cnt;
}
void ins(int x,int y,int a,int b)
{
    add(x,y,b-a);
    d[x]-=a;
    d[y]+=a;
}

int level[N];
bool bfs(int s, int t) {
    memset(level, 0, sizeof level);
    queue<int> q;
    level[s] = 1;
    q.push(s);
    while (!q.empty()) {
        int pos = q.front();q.pop();
        for (int i = head[pos]; i; i = e[i].next) {
            int nx = e[i].to;
            if (!e[i].w || level[nx]) continue;
            level[nx] = level[pos] + 1;
            q.push(nx);
        }
    }
    return level[t];
}
int dfs(int s, int t, int flow) {
    if(s==t||flow==0)return flow;

    int f,ret = 0;
    for (int &i = cur[s],v; i; i = e[i].next) {
         v = e[i].to;
        if (level[v] == level[s] + 1 && (f=dfs(v,t,min(flow,e[i].w)))>0) {
            e[i].w -= f;
            e[i ^ 1].w += f;
            flow -= f;
            ret += f;
            if(!flow)break;
        }
    }
    return ret;
}
int dinic(int s, int t) {
    int ret = 0;
    while (bfs(s, t)) memcpy(cur,head,sizeof cur),ret += dfs(s, t, inf);
    return ret;
}
int n,m,s,t,a,b,c,sum,S,T;
int main()
{   
    s=0;t=5*500+10;
    rep(i,1,500)add(s,i,1),add(i+500,i+2*500,1),add(i+3*500,t,1);
    cin>>n;
    rep(i,1,n)
    {
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        add(x,y+500,inf);add(y+500,y+2*500,inf);add(y+2*500,z+3*500,inf);
    }
    cout<<dinic(s,t);
    return 0;
}

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