HDU3870 Catch the Theves 题解（平面图最小割转最短路）

 

Catch the Theves

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65768/32768 K (Java/Others)
Total Submission(s): 313 Accepted Submission(s): 105

Problem Description

A group of thieves is approaching a museum in the country of zjsxzy,now they are in city A,and the museum is in city B,where keeps many broken legs of zjsxzy.Luckily,GW learned the conspiracy when he is watching stars and told it to zjsxzy.
Zjsxzy decided to caught these thieves,and he let the police to do this,the police try to catch them on their way from A to B. Although the thieves might travel this way by more than one group, zjsxzy's excellent police has already gather the statistics that the cost needed on each road to guard it.
Now ,zjsxzy's conutry can be described as a N*N matrix A,Aij indicates the city(i,j) have bidirectionals road to city(i+1,j) and city(i,j+1),gurad anyone of them costs Aij.
Now give you the map,help zjsxzy to calculate the minimium cost.We assume thieves may travel in any way,and we will catch all passing thieves on a road if we guard it.

Input

The first line is an integer T,followed by T test cases.
In each test case,the first line contains a number N(1<N<=400).
The following N lines,each line is N numbers,the jth number of the ith line is Aij.
The city A is always located on (1,1) and the city B is always located on (n,n).
Of course,the city (i,j) at the last row or last line won't have road to (i,j+1) or (i+1,j).

Output

For each case,print a line with a number indicating the minimium cost to arrest all thieves.

Sample Input

  

   1
3
10 5 5
6 6 20
4 7 9
  

Sample Output

  

   18


   

    

     Hint
    
The map is like this:
    
     
     
   
    
  

Source

2011 Multi-University Training Contest 4 - Host by SDU

 

 

具体见国家集训队2008论文周冬《两极相通——浅析最大—最小定理在信息学竞赛中的应用》

将s-t连一条边，新增一个s面，最外面的为t面，每一个面当做新图的一个顶点。除了s-t面，相邻两个面连一条边，求s-t的最短路即是原图的最小割。

 

code：

/*
平面图最小割转换为最短路
s-t连边增加一个s平面，除了s-t平面，其它平面有相邻则连边。
*/
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
#define N 400*400
using namespace std;

int t,n,a[405][405];
int v,e,low[N],p[N],q[N],adj[N];
bool f[N];
struct Edge
{
	int v,w,next;
}edge[4*N];
void insert(int u,int v,int w)
{
	edge[e].v=v;edge[e].w=w;
	edge[e].next=adj[u];
	adj[u]=e++;
}
int spfa(int s)
{
	memset(f,0,sizeof(f));
	memset(low,0x7f,sizeof(low));
	//fill(low,low+n*n+5,999999999);
	int i,j,k,x,h=0,t=1;
	v=n*n+1;
	q[t]=s;
	low[s]=0;
	while(h!=t)
	{
		h=(h+1)%(v+1);
		x=q[h];
		f[x]=0;
		for(k=adj[x];k!=-1;k=edge[k].next)
		{
			i=edge[k].v;
			j=edge[k].w;
			if(j+low[x]<low[i])
			{
				low[i]=low[x]+j;
				if(!f[i])
				{
					f[i]=1;
					t=(t+1)%(v+1);
					q[t]=i;
				}
			}
		}
	}
	return low[n*n+1];
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		memset(adj,-1,sizeof(adj));
		int i,j;
		e=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				scanf("%d",&a[i][j]);
		n--;
		for(i=0;i<n;i++)
		{
			insert(0,i*n+1,a[i][0]);
			insert(0,(n-1)*n+i+1,a[n][i]);//s平面
			insert(i+1,n*n+1,a[0][i]);
			insert(i*n+n,n*n+1,a[i][n]);//t平面
			for(j=0;j<n;j++)
			{
				if(j+1<n)
				{
					insert(i*n+j+1,i*n+j+2,a[i][j+1]);
					insert(i*n+j+2,i*n+j+1,a[i][j+1]);//相邻平面
				}
				if(i+1<n)
				{
					insert(i*n+j+1,(i+1)*n+j+1,a[i+1][j]);
					insert((i+1)*n+j+1,i*n+j+1,a[i+1][j]);//相邻平面
				}
			}
		}
		printf("%d\n",spfa(0));
	}
}