Bessie and the rest of Farmer John's cows are taking a trip this winter to go skiing. One day Bessie finds herself at the top left corner of an R (1 <= R <= 100) by C (1 <= C <= 100) grid of elevations E (-25 <= E <= 25). In order to join FJ and the other cows at a discow party, she must get down to the bottom right corner as quickly as she can by travelling only north, south, east, and west.
Bessie starts out travelling at a initial speed V (1 <= V <= 1,000,000). She has discovered a remarkable relationship between her speed and her elevation change. When Bessie moves from a location of height A to an adjacent location of eight B, her speed is multiplied by the number 2^(A-B). The time it takes Bessie to travel from a location to an adjacent location is the reciprocal of her speed when she is at the first location.
Find the both smallest amount of time it will take Bessie to join her cow friends.
Bessie starts out travelling at a initial speed V (1 <= V <= 1,000,000). She has discovered a remarkable relationship between her speed and her elevation change. When Bessie moves from a location of height A to an adjacent location of eight B, her speed is multiplied by the number 2^(A-B). The time it takes Bessie to travel from a location to an adjacent location is the reciprocal of her speed when she is at the first location.
Find the both smallest amount of time it will take Bessie to join her cow friends.
Input
* Line 1: Three space-separated integers: V, R, and C, which respectively represent Bessie's initial velocity and the number of rows and columns in the grid.
* Lines 2..R+1: C integers representing the elevation E of the corresponding location on the grid.
* Lines 2..R+1: C integers representing the elevation E of the corresponding location on the grid.
Output
A single number value, printed to two exactly decimal places: the minimum amount of time that Bessie can take to reach the bottom right corner of the grid.
Sample Input
1 3 3 1 5 3 6 3 5 2 4 3
Sample Output
29.00
Hint
Bessie's best route is:
Start at 1,1 time 0 speed 1
East to 1,2 time 1 speed 1/16
South to 2,2 time 17 speed 1/4
South to 3,2 time 21 speed 1/8
East to 3,3 time 29 speed 1/4
Start at 1,1 time 0 speed 1
East to 1,2 time 1 speed 1/16
South to 2,2 time 17 speed 1/4
South to 3,2 time 21 speed 1/8
East to 3,3 time 29 speed 1/4
The question then will use vector T
Here T is the code of the vector is:
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<cmath> #define Inf 0x3f3f3f3f const int maxn=1e5+5; typedef long long ll; using namespace std; ll ksm(ll x,ll y) { ll ans=1; while(y) { if(y&1) { ans=ans*x; } x*=x; y>>=1; } return ans; } struct node { int to; double w; bool friend operator < (node x,node y) { return x.w>y.w; } }; int Map[10005]; vector<node>vec[10005]; double dis[10005]; int vis[10005]; int V,R,C; int dir[4][2]={{0,1},{0,-1},{-1,0},{1,0}}; bool check(int x,int y) { if(x>=1&&x<=R&&y>=1&&y<=C) { return true; } else { return false; } } void init() { for(int t=1;t<=R*C;t++) { dis[t]=100000000000; } } void Dijkstra(int s) { node st; st.to=s; st.w=0; priority_queue<node>q; q.push(st); dis[s]=0; while(!q.empty()) { node now=q.top(); q.pop(); if(vis[now.to])continue; vis[now.to]=1; int len=vec[now.to].size(); for(int t=0;t<len;t++) { node tto=vec[now.to][t]; if(vis[tto.to]==0&&tto.w+dis[now.to]<dis[tto.to]) { tto.w=tto.w+dis[now.to]; dis[tto.to]=tto.w; q.push(tto); } } } } int main() { // std::ios::sync_with_stdio(false); scanf("%d%d%d",&V,&R,&C); init(); for(int t=1;t<=R;t++) { for(int j=1;j<=C;j++) { scanf("%d",&Map[(t-1)*C+j]); } } for(int t=1;t<=R;t++) { for(int j=1;j<=C;j++) { for(int k=0;k<4;k++) { int xx=t+dir[k][0]; int yy=j+dir[k][1]; if(check(xx,yy)) { node s; s.to=(xx-1)*C+yy; s.w=1.0/V*ksm(2 , the Map [(T- . 1 ) * C + J] -Map [ . 1 ]); VEC [(T - . 1 ) * C + J] .push_back (S); } } } } the Dijkstra ( . 1 ); the printf ( " .2f% \ n- " , DIS [R & lt * C]); return 0 ; }
the AC is the adjacency list
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<cmath> const int maxn=1e4+5; typedef long long ll; using namespace std; struct edge { int u,v; double w; int next; }edge[maxn*10]; struct node { int pos; double w; node(int x,double y) { pos=x; w=y; } bool friend operator <(node x,node y) { return x.w>y.w; } }; int n,m,s,x,y,z,tot = 0,V; bool check(int x,int y) { if(x>=1&&x<=n&&y>=1&&y<=m) { return true; } else { return false; } } int head[10005]; double d[10005]; int vis[10005]; int a[10005]; int dist[4][2] = {{-1,0},{0,-1},{1,0},{0,1}}; void add(int u,int v,double w) { edge[++tot].u=u; edge[tot].v=v; edge[tot].w=w; edge[tot].next=head[u]; head[u]=tot; return ; } void Dijkstra(int s) { priority_queue<node>q; d[s]=0; q.push(node(s,0)); while(!q.empty()) { node now=q.top(); q.pop(); //cout<<now.pos<<endl; if(vis[now.pos])continue; vis[now.pos]=1; for(int i=head[now.pos];i!=-1;i=edge[i].next) { int ne=edge[i].v; double ww=edge[i].w; if(d[now.pos]+ww<d[ne]) { d[ne]=d[now.pos]+ww; q.push(node(ne,d[ne])); } } } return ; } int main() { scanf("%d%d%d",&V,&n,&m); memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); for(int t=1;t<=n*m;t++) { d[t]=1000000000000; } for(int i = 1;i <= n; ++i) for(int j = 1;j <= m; ++j) scanf("%d",&a[(i-1)*m+j]); for(int i = 1;i <= n; ++i) for(int j = 1;j <= m; ++j) for(int k = 0;k < 4; ++k) { int x = i + dist[k][0]; int y = j + dist[k][1]; if(check(x,y)) { double v = 1.0 / V * pow(2.0 , a[(i-1)*m+j] - a[1]); add((i-1)*m+j,(x-1)*m+y,v); } } Dijkstra(1); printf("%.2f\n",d[n * m]); return 0; }