BZOJ 2709 Violet 1 迷宫花园

2709: [Violet 1]迷宫花园

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 976  Solved: 340
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Description

 

Input

Output

Sample Input

5
10.28 9 9
#########
# #
# # # # #
#S# #
##### # #
## # #
# ### ###
##E #
#########
4.67 9 9
#########
# ## ##
### #S# #
# # E ##
# # #####
# ## ###
# ##### #
# # #
#########
39.06 9 9
#########
# #
# # # # #
# E# # #
# # # # #
## ### #
# # #S# #
##### #
#########
24.00 9 9
#########
# ##
# # ## ##
# # ##
###S## E#
### # ##
# # # #
##### # #
#########
25.28 9 9
#########
# S##E# #
# ### # #
# ## #
# ## ###
# # ####
# # # ###
# #
#########

Sample Output

0.41000
4.67000
3.34000
5.00000
1.69000

HINT

 

 

Source

POJ 3897 Maze Stretching

实数二分+最短路判定

很水的一道题，注意读数据的时候，空格scanf和cin会自动忽略，用gets去读

#include <bits/stdc++.h>
#define ll long long
#define eps 1e-7
using namespace std;
inline int read(){
    int x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int MAXN=1e4+10;
struct node{
    int x,y;
}e[MAXN];
const int dx[4]={0,0,-1,1};
const int dy[4]={-1,1,0,0};
int n,m,T,vis[MAXN],S,E;
char ch[110][110];
double dis[MAXN],L;
typedef pair < double,int > pii;
priority_queue < pii,vector<pii>,greater<pii> > q;
namespace zhangenming{
    inline int di(int xx,int yy){
        return (xx-1)*m+yy;
    }
    inline void dijsktra(double vv){
        for(int i=1;i<=n*m;i++){
            dis[i]=1e30;
        }
        memset(vis,0,sizeof(vis));
        while(!q.empty()) q.pop(); 
        q.push(make_pair(0,S));dis[S]=0;
        while(!q.empty()){
            int tn=q.top().second;q.pop();
            if(vis[tn]) continue;vis[tn]=1;
            for(int i=0;i<4;i++){
                if(i>=2){
                    int xx=e[tn].x+dx[i];int yy=e[tn].y+dy[i];
                    if(yy==0||yy==m+1||ch[xx][yy]=='#') continue;
                    if(dis[di(xx,yy)]>dis[tn]+vv){
                        dis[di(xx,yy)]=dis[tn]+vv;
                        q.push(make_pair(dis[di(xx,yy)],di(xx,yy)));
                    }
                }
                else{
                    int xx=e[tn].x+dx[i];int yy=e[tn].y+dy[i];
                    if(xx==0||xx==n+1||ch[xx][yy]=='#') continue;
                    if(dis[di(xx,yy)]>dis[tn]+1){
                        dis[di(xx,yy)]=dis[tn]+1;
                        q.push(make_pair(dis[di(xx,yy)],di(xx,yy)));
                    }
                }
            }
        }
    }
    void work(){
        double l=0;double r=10;
        while((r-l)>eps){
            double mid=(l+r)*0.5;
            dijsktra(mid);
            if((L-dis[E])>-eps) l=mid;
            else r=mid;
        }
        dijsktra(r);
        if(dis[E]-L<=eps) printf("%.5lf\n",r);
        else printf("%.5lf\n",l);
    }
    void solve(){
        T=read();
        while(T--){
            scanf("%lf",&L);n=read();m=read();char c=getchar();
            for(int i=1;i<=n;i++){
                gets(ch[i]+1);
                for(int j=1;j<=m;j++){
                    if(ch[i][j]=='S') S=di(i,j);
                    if(ch[i][j]=='E') E=di(i,j);
                }
            }
            for(int i=1;i<=n;i++){
                for(int j=1;j<=m;j++){
                    e[(i-1)*m+j].x=i;e[(i-1)*m+j].y=j;
                }
            }
            work();
        }
    }
}
int main(){
    using namespace zhangenming;
    solve();
    return 0;
}