Xiao Ming climbing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1227 Accepted Submission(s): 344
Problem Description
Due to the curse made by the devil,Xiao Ming is stranded on a mountain and can hardly escape.
This mountain is pretty strange that its underside is a rectangle which size is
n∗m and every little part has a special coordinate
(x,y)and a height
H.
In order to escape from this mountain,Ming needs to find out the devil and beat it to clean up the curse.
At the biginning Xiao Ming has a fighting will
k,if it turned to
0 Xiao Ming won't be able to fight with the devil,that means failure.
Ming can go to next position
(N,E,S,W)from his current position that time every step,
(abs(H1−H2))/k 's physical power is spent,and then it cost
1 point of will.
Because of the devil's strong,Ming has to find a way cost least physical power to defeat the devil.
Can you help Xiao Ming to calculate the least physical power he need to consume.
Input
T(T≤10), indicating the number of testcases.
Then
T testcases follow.
The first line contains three integers
n,m,k ,meaning as in the title
(1≤n,m≤50,0≤k≤50).
Then the
N ×
M matrix follows.
In matrix , the integer
H meaning the height of
(i,j),and '#' meaning barrier (Xiao Ming can't come to this) .
Then follow two lines,meaning Xiao Ming's coordinate
(x1,y1) and the devil's coordinate
(x2,y2),coordinates is not a barrier.
Output
NoAnswer" otherwise.
(The result should be rounded to 2 decimal places)
Sample Input
3 4 4 5 2134 2#23 2#22 2221 1 1 3 3 4 4 7 2134 2#23 2#22 2221 1 1 3 3 4 4 50 2#34 2#23 2#22 2#21 1 1 3 3
Sample Output
1.03 0.00 No Answer
Source
这个题真是快折磨死自己了，各种WA。
一道迷宫题目：
只让你走k步，每一步都消耗abs(h1-h2)/k的体力，问能否从起点到达终点，并且消耗体力最少！
思路：
BFS可以用dp[x][y][k]来表示状态。表示走到x，y这个点时，还有k步可走时，的体力消耗总值。
建立一个结构体包含，int x,y,k doulbe will， 这里will记录体力消耗总值。
构建一个优先队列即可，优先队列中优先级高的是will高的那个结构体。因为体力消耗是越来越多嘛，所以越后面的就以后面的为基础在继续找。
判断k <= 0时就不停的continue，这样就可以把k = 0 考虑在内了！
接下来就是正常的bfs
代码如下：
#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
const int maxn = 50 + 10;
char mep[maxn][maxn];
int n,m,k,gx,gy,sx,sy;
double dp[maxn][maxn][maxn];
bool vis[maxn][maxn][maxn];
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
const double INF = 1e18;
const double eps = 1e-8;
struct Node{
int x,y,k;
double will;
Node(int x,int y,int k,double will):x(x),y(y),k(k),will(will){}
bool operator < (const Node&a)const {
return will > a.will;
}
};
bool init(int x,int y){
return x > 0 && x <= n && y > 0 && y <= m && mep[x][y] != '#';
}
void bfs(){
memset(vis,false,sizeof(vis));
priority_queue<Node>q;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j)
for (int h = 0; h <=k; ++h)
dp[i][j][h] = INF;
q.push(Node(sx,sy,k,0));
dp[sx][sy][k]=0;
while(!q.empty()){
Node u = q.top();q.pop();
if (vis[u.x][u.y][u.k])continue;
vis[u.x][u.y][u.k]=true;
for (int i = 0; i < 4; ++i){
Node v(u.x+dx[i],u.y+dy[i],u.k-1,u.will);
if (v.k <= 0 || !init(v.x,v.y))continue;
v.will+=abs(mep[u.x][u.y] - mep[v.x][v.y])*1.0 / u.k;
if (dp[v.x][v.y][v.k] > v.will){
dp[v.x][v.y][v.k] = v.will;
q.push(v);
}
}
}
}
int main()
{
int T;
cin >> T;
while(T--){
cin >> n >> m >> k;
for (int i = 1; i <= n; ++i)scanf("%s",mep[i] + 1);
cin >> sx >> sy >> gx >> gy;
if (!k || abs(abs(sx-gx) + abs(gx-sy)) > k){printf("No Answer\n");continue;}
bfs();
double ans = INF;
for (int i = 1; i <= k; ++i)ans = min(ans,dp[gx][gy][i]);
if (fabs(ans - INF) < eps)printf("No Answer\n");
else printf("%.2lf\n",ans);
}
return 0;
}
