UVA10967 The Great Escape(最短路)

题意:

要求从左下角到右上角,中间会经过旋转门,而旋转门的房间固定只有一个房间能进入.旋转门每转90度,会消耗时间d,然而每个旋转门所消耗的时间都不同.而一般进入房间消耗单位1时间,求最少时间!

分析:其实是个很简单的最短路的题,,,,只是弱菜的我题意看错,每个旋转门消耗时间不同....还有写代码的能力不强,...老写错和漏掉什么的!唉~

只要简单处理下旋转门即可!

// File Name: 10967.cpp
// Author: Zlbing
// Created Time: 2013/5/29 10:18:14

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int MAXN=205;
int n,m,T;
char G[MAXN][MAXN];
int D[MAXN][MAXN];
int dx[4]= {0,0,1,-1};
int dy[4]= {1,-1,0,0};
int vis[MAXN][MAXN];
struct node
{
    int x,y,cost;
    bool operator <(const node& rsh)const
    {
        return cost>rsh.cost;
    }
};
int find(char a)
{
    if(a=='W')return 1;
    if(a=='N')return 2;
    if(a=='E')return 3;
    if(a=='S')return 4;
    if(a=='.')return 0;
    if(a=='#')return -1;
}
int cha(int ax,int ay,int bx,int by)
{
    if(ax==bx&&by>ay)return 1;
    if(ay==by&&ax<bx)return 2;
    if(ax==bx&&ay>by)return 3;
    if(ay==by&&ax>bx)return 4;
}
int cha2(int a,int b)
{
    if(b==1){
        b=3;
    }else
    if(b==2){
        b=4;
    }else
    if(b==3)b=1;
    else
    if(b==4)b=2;
    if(abs(a-b)<=2)return abs(a-b);
    if(abs(a-b)>2)return 1;
}
int dfs()
{
    priority_queue<node> Q;
    CL(vis,-1);
    Q.push((node)
    {
        n-1,0,0
    });
    node t,tt;
    vis[n-1][0]=0;
    while(!Q.empty())
    {
        t=Q.top();
        Q.pop();
      //  printf("now---x=%d y=%d cost=%d\n",t.x,t.y,t.cost);
        if(t.x==0&&t.y==m-1)return t.cost;
        for(int i=0; i<4; i++)
        {
            int xx=t.x+dx[i];
            int yy=t.y+dy[i];
            if(xx<0||xx>n-1||yy<0||yy>m-1)continue;
            int to=find(G[xx][yy]);
            int too=cha(t.x,t.y,xx,yy);
            if(find(G[t.x][t.y])==0&&to==0)
            {
                if(vis[xx][yy]==-1||t.cost+1<vis[xx][yy])
                {
                    vis[xx][yy]=t.cost+1;
                 //   printf("1pushinto---x=%d y=%d cost=%d\n",xx,yy,t.cost+1);
                    Q.push((node)
                    {
                        xx,yy,t.cost+1
                    });
                }
            }
            else if(too==to||to==0)
            {
                if(find(G[t.x][t.y])==0)
                {
                    if(vis[xx][yy]==-1||t.cost+1<vis[xx][yy])
                    {
                        vis[xx][yy]=t.cost+1;
                    //    printf("2pushinto---x=%d y=%d cost=%d\n",xx,yy,vis[xx][yy]);
                        Q.push((node)
                        {
                            xx,yy,t.cost+1
                        });
                    }
                }
                else
                {
                    int c=cha2(find(G[t.x][t.y]),too);
                //    printf("too=%d c=%d\n",too,c);
                    if(vis[xx][yy]==-1||t.cost+1+D[t.x][t.y]*c<vis[xx][yy])
                    {
                        vis[xx][yy]=t.cost+1+D[t.x][t.y]*c;
                 //       printf("3pushinto---x=%d y=%d cost=%d\n",xx,yy,vis[xx][yy]);
                        Q.push((node)
                        {
                            xx,yy,t.cost+1+D[t.x][t.y]*c
                        });
                    }
                }
            }
            else continue;
        }
    }
    return vis[0][m-1];
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        CL(D,0);
        REP(i,0,n-1)
        scanf("%s",G[i]);
        REP(i,0,n-1)
        REP(j,0,m-1)
        {
            if(G[i][j]!='.'&&G[i][j]!='#')
                scanf("%d",&D[i][j]);
        }
        int ans=dfs();
        if(ans==-1)printf("Poor Kianoosh\n");
        else printf("%d\n",ans);
    }
    return 0;
}