【BZOJ2346】Lamp

题目描述

Casper is designing an electronic circuit on a N \times MN×M rectangular grid plate. There are N \times MN×M square tiles that are aligned to the grid on the plate. Two (out of four) opposite corners of each tile are connected by a wire.

A power source is connected to the top left corner of the plate. A lamp is connected to the bottom right corner of the plate. The lamp is on only if there is a path of wires connecting power source to lamp. In order to switch the lamp on, any number of tiles can be turned by 90° (in both directions).

In the picture above the lamp is off. If any one of the tiles in the second column from the right is turned by 90° , power source and lamp get connected, and the lamp is on.

Write a program to find out the minimal number of tiles that have to be turned by 90° to switch the lamp on.

输入格式：

The first line of input contains two integer numbers NN and MM, the dimensions of the plate. In each of the following NN lines there are MM symbols – either \ or / – which indicate the direction of the wire connecting the opposite vertices of the corresponding tile.

输出格式：

There must be exactly one line of output. If it is possible to switch the lamp on, this line must contain only one integer number: the minimal number of tiles that have to be turned to switch on the lamp. If it is not possible, output the string: NO SOLUTION

输入样例：

3 5
\\/\\
\\///
/\\\\

输出样例：

1

说明

对于所有数据，

 

解析：

       对角线建边跑最短路即可。

 

代码：

#include 
using namespace std;

const int inf=1e9;
const int Max=2000010;
int n,m,size;
int first[251005],dis[251005],vis[251005];
char d[505][505];
struct shu{int next,to,len;};
shu edge[Max];

inline int get_int()
{
    int x=0,f=1;
    char c;
    for(c=getchar();(!isdigit(c))&&(c!='-');c=getchar());
    if(c=='-') f=-1,c=getchar();
    for(;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+c-'0';
    return x*f;
}

inline void build(int x,int y,int z)
{
    edge[++size].next=first[x];
    first[x]=size;
    edge[size].to=y,edge[size].len=z;
}

inline void dijkstra()
{
    priority_queue >q;
    for(int i=0;i<=n*m;i++) dis[i]=inf;
    dis[1]=0,q.push(make_pair(0,1));
    while(!q.empty())
    {
      int point=q.top().second;q.pop();
      if(vis[point])continue;vis[point]=1;
      for(int u=first[point];u;u=edge[u].next)
      {
      	int to=edge[u].to;
      	if(dis[to] > dis[point] + edge[u].len)
      	{
      	  dis[to] = dis[point] + edge[u].len;
      	  q.push(make_pair(-dis[to],to));
      	}
      }
    }
}

int main()
{
    n=get_int(),m=get_int();
    for(int i=1;i<=n;i++)
    {
      for(int j=1;j<=m;j++) scanf("%c",&d[i][j]);
      scanf("\n");
    }
    n++,m++;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=m;j++)
      {	  	
        if(j!=m)
        {
          if(i!=1) build((i-1)*m+j,(i-2)*m+j+1,d[i-1][j]=='/' ? 0 : 1);
          if(i!=n) build((i-1)*m+j,i*m+j+1,d[i][j]=='/' ? 1 : 0);
        }
        if(j!=1)
        {
      	  if(i!=1) build((i-1)*m+j,(i-2)*m+j-1,d[i-1][j-1]=='/' ? 1 : 0);
          if(i!=n) build((i-1)*m+j,i*m+j-1,d[i][j-1]=='/' ? 0 : 1);
        }
      }

    dijkstra();
    if(dis[n*m]==dis[0]) cout<<"NO SOLUTION";
    else cout<