NOI 2010 海拔 ——平面图转对偶图

Posted 2020-08-24

【题目分析】

    可以知道，所有的海拔是0或1

    最小割转最短路，就可以啦

    SPFA被卡，只能换DIJ

【代码】

#include <cstdio>
#include <cstring>

#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

#define maxn 2000005

int h[maxn],to[maxn],ne[maxn],w[maxn],en=0;

int read()
{
    int x=0,f=1; char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) f=-1; ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}

int dis[maxn];
struct cmp{
    bool operator () (const int &a,const int &b)
    {return dis[a]>dis[b];}
};

struct Table
{
    int h[maxn],ne[maxn],to[maxn],w[maxn],vis[maxn];
    priority_queue <int,vector <int>,cmp> q;
    int en;
    void init()
    {
        memset(h,-1,sizeof h);
        en=0;
        while (!q.empty()) q.pop();
    }
    void add(int a,int b,int c)
    {
        ne[en]=h[a];
        to[en]=b;
        w[en]=c;
        h[a]=en++;
    }
    void dij(int s)
    {
        memset(dis,0x3f,sizeof dis);
        memset(vis,0,sizeof vis);
        dis[s]=0;
        q.push(s);
        vis[s]=1;
        while (!q.empty())
        {
            int x=q.top(); q.pop(); vis[x]=0;
            for (int i=h[x];i>=0;i=ne[i])
            {
                if (dis[to[i]]>dis[x]+w[i])
                {
                    dis[to[i]]=dis[x]+w[i];
                    if (!vis[to[i]])
                    {
                        vis[to[i]]=1;
                        q.push(to[i]);
                    }
                }
            }
        }
    }
}map;

void add(int a,int b,int c)
{
	to[en]=b;
	ne[en]=h[a];
	w[en]=c;
	h[a]=en++;
}

int n,S=0,T=maxn-1;
int hash[605][605],cnt=0;
int inq[maxn];

void SPFA()
{
	queue <int> q;
	memset(dis,0x3f,sizeof dis);
	while (!q.empty()) q.pop();
	q.push(S); inq[S]=1; dis[S]=0;
	while (!q.empty())
	{
		int x=q.front(); q.pop(); inq[x]=0;
		for (int i=h[x];i>=0;i=ne[i])
		{
			if (dis[to[i]]>dis[x]+w[i])
			{
				dis[to[i]]=dis[x]+w[i];
				if (!inq[to[i]])
				{
					inq[to[i]]=1;
					q.push(to[i]);
				}
			}
		}
	}
	printf("%d\n",dis[T]);
}

int main()
{
	map.init();
	memset(h,-1,sizeof h);
	n=read();
	for (int i=0;i<=n+1;++i)
		for (int j=0;j<=n+1;++j)
			hash[i][j]=++cnt;
	for (int i=0;i<=n+1;++i) hash[0][i]=maxn-1;
	for (int i=0;i<=n+1;++i) hash[n+1][i]=0;
	for (int i=0;i<=n+1;++i) hash[i][0]=0;
	for (int i=0;i<=n+1;++i) hash[i][n+1]=maxn-1;
	for (int i=1;i<=n+1;++i)
		for (int j=1;j<=n;++j)
		{
			int tmp=read();
			map.add(hash[i][j],hash[i-1][j],tmp);
		}
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n+1;++j)
		{
			int tmp=read();
			map.add(hash[i][j-1],hash[i][j],tmp);
		}
	for (int i=1;i<=n+1;++i)
		for (int j=1;j<=n;++j)
		{
			int tmp=read();
			map.add(hash[i-1][j],hash[i][j],tmp);
		}
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n+1;++j)
		{
			int tmp=read();
			map.add(hash[i][j],hash[i][j-1],tmp);
		}
	map.dij(S);
	printf("%d\n",dis[T]);
}