洛谷2046 NOI2010海拔

QwQ题目太长这里就不复制了

这个题…算是个比较经典的平面图最小割变成对偶图的最短路了QwQ

首先考虑最小割应该怎么做。

有一个性质，就是每个点的海拔要么是1，要么是0

QwQ不过这个我不会证明啊

那么既然知道了这个性质，我们对于地图上的每个点，实际上就是划分成两个集合，一个是 $1$ ，一个是 $0$

那么直接最小割就行了

// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>

using namespace std;

inline int read()
{
   int x=0,f=1;char ch=getchar();
   while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}
   while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
   return x*f;
}

const int maxn = 110*110;
const int maxm = 1e6+1e2;
const int inf = 1e9;

int point[maxn];
int nxt[maxm],to[maxm],val[maxm];
int cnt=1;
int h[maxn];
queue<int> q;
int n,m;
int s,t;

void addedge(int x,int y,int w)
{
    nxt[++cnt]=point[x];
    to[cnt]=y;
    val[cnt]=w;
    point[x]=cnt;
}

void insert(int x,int y,int w)
{
    addedge(x,y,w);
    addedge(y,x,0);
}

bool bfs(int s)
{
    memset(h,-1,sizeof(h));
    h[s]=0;
    q.push(s);
    while (!q.empty())
    {
        int x  = q.front();
        q.pop();
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (val[i]>0 && h[p]==-1)
            {
                h[p]=h[x]+1;
                q.push(p);
            }
        }
    }
    if (h[t]==-1) return false;
    else return true;
}

int dfs(int x,int low)
{
    if (x==t || low==0) return low;
    int totflow=0;
    for (int i=point[x];i;i=nxt[i])
    {
        int p = to[i];
        if (val[i]>0 && h[p]==h[x]+1)
        {
            int tmp = dfs(p,min(low,val[i]));
            low-=tmp;
            totflow+=tmp;
            val[i]-=tmp;
            val[i^1]+=tmp;
            if (low==0) return totflow;
        }
    }
    if (low>0) h[x]=-1;
    return totflow;
}

int dinic()
{
    int ans=0;
    while (bfs(s))
    {
        ans=ans+dfs(s,inf);
    }
    return ans;
}


int main()
{
  n=read();
  n++;
  s=1;
  t=n*n;
  for (int i=1;i<=n;i++)
  {
    //int now =(i-1)*n;
    for (int j=1;j<n;j++)
    {
        int x = read();
        //cout<<x<<endl;
        insert((i-1)*n+j,(i-1)*n+j+1,x);
        //cout<<(i-1)*n+j<<" "<<(i-1)*n+j+1<<endl; 
    }
  }
  for (int i=1;i<n;i++)
    for (int j=1;j<=n;j++)
    {
        int x = read();
        insert((i-1)*n+j,i*n+j,x);
    }
  for (int i=1;i<=n;i++)
  {
    for (int j=1;j<n;j++)
    {
        int x = read();
        insert((i-1)*n+j+1,(i-1)*n+j,x);    
    }
  }
  for (int i=1;i<n;i++)
    for (int j=1;j<=n;j++)
    {
        int x = read();
        insert(i*n+j,(i-1)*n+j,x);
    }
  cout<<dinic()<<endl;
  return 0;
}

不过这个最小割的复杂度是爆炸的，显然没法通过这个题，那么我们这时候就需要用到一个很关键的性质了

平面图最小割等于对偶图的最短路

那么什么是对偶图呢？
简单来说，就是把原图的每个封闭面，看成一个点，然后原图的每一种割，对应着新图 $s到t$ 的一条路径

但是QwQ这里先留跟个坑，就是关于边的方向的问题….这里还不是很理解呢

转化成新图，建好图之后，直接从 $S$ 开始跑最短路， $dis[t]$ 就是答案
一般原图的st和新图的st成对角线的关系

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define pa pair<long long,long long>
#include<queue>
using namespace std;

inline long long read()
{
   long long x=0,f=1;char ch=getchar();
   while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}
   while (isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
   return x*f;
}

const int maxn = 510;
const int N = maxn*maxn;
const int maxm = 2e6+1e2;

int a[maxn][maxn][maxn];
int point[N],nxt[maxm],to[maxm];
int cnt;
int vis[N];
int n,m;
long long dis[N],val[maxm];
priority_queue<pa,vector<pa>,greater<pa> > q;
int s,t;

void addedge(int x,int y,long long w){
   nxt[++cnt]=point[x];
   to[cnt]=y;
   val[cnt]=w;
   point[x]=cnt;
}

void splay(int s)
{
    memset(vis,0,sizeof(vis));
    memset(dis,127/3,sizeof(dis));
    //cout<<dis[1]<<endl;
    dis[s]=0;
    q.push(make_pair(0,s));
    while (!q.empty())
    {
        //cout<<1<<endl;
        int x = q.top().second;
        q.pop();
        //cout<<x<<endl;
        if (vis[x]) continue;
        vis[x]=1;
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (dis[p]>dis[x]+val[i])
            {
                dis[p]=dis[x]+val[i];
                //cout<<dis[p]<<" "<<p<<endl;
                q.push(make_pair(dis[p],p));
                //cout<<endl;
            }
        }
    }
}

inline int getnum(int x,int y)
{
    if (x==0 || y==n) return t;
    if (x==n || y==0 ) return s;
    return (x-1)*(n-1)+y;
}
int main()
{
  n=read();
  n++;
  s=N-6;
  t=s+1;
  for (int i=1;i<=n;i++)
    for (int j=1;j<n;j++)
    {
        long long x=read();
        addedge(getnum(i,j),getnum(i-1,j),x);
        //cout<<getnum(i,j)<<" "<<getnum(i-1,j)<<endl;
    }
  for (int i=1;i<n;i++)
    for (int j=1;j<=n;j++)
    {
        long long x = read();
        addedge(getnum(i,j-1),getnum(i,j),x);
        //cout<<getnum(i,j-1)<<" "<<getnum(i,j)<<endl;  
        //cout<<x<<endl;
    }
  for (int i=1;i<=n;i++)
    for (int j=1;j<n;j++)
    {
        long long x=read();
        addedge(getnum(i-1,j),getnum(i,j),x);
    }  
   for (int i=1;i<n;i++)
    for (int j=1;j<=n;j++)
    {
        long long x = read();
        addedge(getnum(i,j),getnum(i,j-1),x);   
    }
    splay(s);
    cout<<dis[t];
  return 0;
}

洛谷2046 NOI2010海拔

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