WOJ 1546 Maze

怎么简单的题目，比赛时看都没看。。。。。不知道当时在干什么。。。。
用SublimeText敲代码的手感就是好

Problem 1546 - J - Maze

Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 68 Accepted: 8 Special Judge: No

Description

Get out from the maze!

He looks around, finding that the maze has only one exit but it has been locked. He has not enough energy to break the door and he must move out to find extra bonuses of energy hidden in the maze. He can move toward 4 directions: left, right, up, and down, which will take 1 second in a single move.

Unfortunately, there is an evil machine which will detect whether he is still there or not every L seconds in the initial place where he is located. And if he is out of the maze, the machine will surely be out of service.

Right after he is in the exit grid, he will be considered to be outside the maze.

Input

The input consists of one or more test cases.

First line of each test case contains 5 integers W(1<=W<=50), H(1<=H<=50), L(1<=L<=1,000,000), M(1<=M<=10), and S(0<=S<=10,000,000), indicating:
W: the width of the maze.
H: the height of the maze.
L: time of the round of the machine.
M: the number of the bonuses of energy.
S: energy needed to break the lock.

The next line contains M integers indicating the list of the energy-bonus.

Then there comes the map of the maze:
#: wall
.: empty place
$: initial position
<: exit grid
@: energy bonuses

Energy bonuses are ordered from the top to bottom, left to the right.

Output

For each test case, output “YES” or “NO”

Sample Input

3
4 4 2 2 200
100 200
####
#$@#
#@<#
####

4 4 1 2 300
100 200
####
#$@#
#@<#
####

12 5 13 2 400
100 200
############
#@.........#
#.########.#
#$...@....<#
############

Sample Output

YES
NO
NO

Hint

Source

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>using namespace std;const int INF=0x3f3f3f3f;
const int dir_x[4]={-1,1,0,0};
const int dir_y[4]={0,0,-1,1};typedef pair<int,int> pII;char maze[60][60];
int w,h,l,m,s,egy[20],EG[60][60],step[60][60];
int qi[60][60],zhong[60][60];
vector<pII> egpoint;void init()
{egpoint.clear();memset(maze,0,sizeof(maze));memset(egy,0,sizeof(egy));memset(EG,0,sizeof(EG));memset(step,0,sizeof(step));memset(qi,63,sizeof(qi));memset(zhong,63,sizeof(zhong));
}int main()
{int T_T;scanf("%d",&T_T);
while(T_T--)
{int sum=0;init();scanf("%d%d%d%d%d",&w,&h,&l,&m,&s);    for(int i=0;i<m;i++) scanf("%d",egy+i),sum+=egy[i];for(int i=0;i<h;i++) scanf("%s",maze[i]); //for(int i=0;i<h;i++) cout<<maze[i]<<endl;if(sum<s){puts("NO"); continue;///根本没有足够的能量}pII IP,EXIT;int cur=0;for(int i=0;i<h;i++){for(int j=0;j<w;j++){if(maze[i][j]=='$'){IP=make_pair(i,j);}else if(maze[i][j]=='<'){EXIT=make_pair(i,j);}else if(maze[i][j]=='@'){EG[i][j]=egy[cur++];egpoint.push_back(make_pair(i,j));}}}queue<pII> q;q.push(IP);qi[IP.first][IP.second]=0;while(!q.empty())//从起点到各个点的距离{pII u=q.front(),v; q.pop();for(int i=0;i<4;i++){v.first=u.first+dir_x[i];v.second=u.second+dir_y[i];if(maze[v.first][v.second]=='#'||qi[v.first][v.second]!=INF) continue;qi[v.first][v.second]=qi[u.first][u.second]+1;q.push(v);}}if(qi[EXIT.first][EXIT.second]>l){puts("NO"); continue;  ///即使不要能量也跑不出去}while(!q.empty()) q.pop();q.push(EXIT);zhong[EXIT.first][EXIT.second]=0;while(!q.empty())///各个点到出口的距离{pII u=q.front(),v; q.pop();for(int i=0;i<4;i++){v.first=u.first+dir_x[i];v.second=u.second+dir_y[i];if(maze[v.first][v.second]=='#'||zhong[v.first][v.second]!=INF) continue;zhong[v.first][v.second]=zhong[u.first][u.second]+1;q.push(v);}}int CAN=0;for(int i=0,sz=egpoint.size();i<sz;i++)    {int X=egpoint[i].first;int Y=egpoint[i].second;if(qi[X][Y]*2<=l||zhong[X][Y]*2<=l||qi[X][Y]+zhong[X][Y]<=l) CAN+=EG[X][Y];}if(CAN>=m) puts("YES");else puts("NO");
}return 0;
}

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