BZOJ 2709: [Violet 1]迷宫花园
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
.gif)
1 /* 2 分析：符合二分的原理：当v变大，dis一定变大，而且v的具体范围很小，才是0--10，符合二分原理。 3 二分出一个V，就用spfa求一次最短路，看看最短的长度与L大小关系，以此来二分。 4 */ 5 #include<cmath> 6 #include<iostream> 7 using namespace std; 8 #include<cstdio> 9 #include<queue> 10 #include<cstring> 11 #define R 110 12 int T,r,c; 13 bool inque[R*R]; 14 char ditu[R][R]; 15 double L,z,y; 16 double dist[R*R]; 17 int head[10010],bh=0,sta,endd,bhh[R][R]; 18 struct Edge{ 19 int v,last; 20 double w; 21 }edge[40005]; 22 int t=0; 23 void input() 24 { 25 scanf("%lf%d%d ",&L,&r,&c); 26 for(int i=1;i<=r;++i) 27 { 28 for(int j=1;j<=c;++j) 29 { 30 scanf("%c",&ditu[i][j]); 31 if(ditu[i][j]==32) 32 { 33 bh++; 34 bhh[i][j]=bh; 35 } 36 if(ditu[i][j]=='S')bhh[i][j]=sta=++bh; 37 if(ditu[i][j]=='E')bhh[i][j]=endd=++bh; 38 } 39 scanf(" "); 40 } 41 42 } 43 void add_edge(int i,int j) 44 { 45 if(i-1>0&&ditu[i-1][j]!='#') {++t;edge[t].v=bhh[i-1][j];edge[t].w=-1;edge[t].last=head[bhh[i][j]];head[bhh[i][j]]=t;} 46 if(i<r&&ditu[i+1][j]!='#') {++t;edge[t].v=bhh[i+1][j];edge[t].w=-1;edge[t].last=head[bhh[i][j]];head[bhh[i][j]]=t;} 47 if(j-1>0&&ditu[i][j-1]!='#') {++t;edge[t].v=bhh[i][j-1];edge[t].w=1;edge[t].last=head[bhh[i][j]];head[bhh[i][j]]=t;} 48 if(j<c&&ditu[i][j+1]!='#') {++t;edge[t].v=bhh[i][j+1];edge[t].w=1;edge[t].last=head[bhh[i][j]];head[bhh[i][j]]=t;} 49 } 50 void build_tu() 51 { 52 for(int i=1;i<=r;++i) 53 for(int j=1;j<=c;++j) 54 if(ditu[i][j]!='#') 55 { 56 add_edge(i,j); 57 } 58 } 59 double SPFA(double ww) 60 { 61 for(int i=1;i<=bh;++i) 62 dist[i]=(1<<30)-1; 63 dist[sta]=0; 64 memset(inque,false,sizeof(inque)); 65 queue<int>Q; 66 Q.push(sta); 67 inque[sta]=true; 68 while(!Q.empty()) 69 { 70 int nowt=Q.front(); 71 Q.pop(); 72 inque[nowt]=false; 73 for(int l=head[nowt];l;l=edge[l].last) 74 { 75 if(edge[l].w<0) 76 { 77 if(dist[edge[l].v]>dist[nowt]+ww) 78 { 79 dist[edge[l].v]=dist[nowt]+ww; 80 if(!inque[edge[l].v]) 81 { 82 inque[edge[l].v]=true; 83 Q.push(edge[l].v); 84 } 85 } 86 } 87 else { 88 if(dist[edge[l].v]>dist[nowt]+edge[l].w) 89 { 90 dist[edge[l].v]=dist[nowt]+edge[l].w; 91 if(!inque[edge[l].v]) 92 { 93 inque[edge[l].v]=true; 94 Q.push(edge[l].v); 95 } 96 } 97 } 98 } 99 } 100 return dist[endd]; 101 } 102 int main() 103 { 104 cin>>T; 105 while(T--) 106 { 107 input(); 108 build_tu(); 109 z=0;y=10; 110 while(z<=y) 111 { 112 double mid=(z+y)/2; 113 double ans=SPFA(mid); 114 if(ans>=L) y=mid-0.000001;/*注意这里要加0.000001，之前的二分加1，是为了去一个区间（int），但是现在是double，所以要+0 115 .000001。*/ 116 else z=mid+0.000001; 117 } 118 printf("%0.5lf ",y); 119 } 120 121 return 0; 122 }