程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> hdu1242

hdu1242

編輯：C++入門知識

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14614 Accepted Submission(s): 5300

Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that approach Angel is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. . stands for road, a stands for Angel, and r stands for each of Angel's friend.

Process to the end of the file.

Output For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing Poor ANGEL has to stay in the prison all his life.

Sample Input

7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........

Sample Output

Author CHEN, Xue
Source ZOJ Monthly, October 2003
Recommend Eddy | We have carefully selected several similar problems for you: 1240 1072 1241 1253 1372 這個題目最開始是沒有說清楚。。請看。。

Angel's friends

結果單組就可以過了。。。還有就是這個題目因為碰到守衛會出現時間的不同變化。。所以我運用了剪枝。。。如果步數大於已經確定的步數，則不需要入隊。。這個還是挺高效的。。0ms。。。。代碼如下：

#include
#include
#include
#define INF 0x3f3f3f3f
const int maxn=200+10;
using namespace std;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
int step[maxn][maxn];
int n,m;
int sum;
char map[maxn][maxn];

struct Map
{
    int x,y;
    int time;
};
queueq;
Map start,end;

int check(int x,int y)
{
    if(x>=1&&x<=n&&y>=1&&y<=m)
      return 1;
    return 0;
}

int bfs()
{
    Map current,next;
    step[start.x][start.y]=0;
    q.push(start);
    while(!q.empty())
    {
        current=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            next.x=current.x+dx[i];
            next.y=current.y+dy[i];
            next.time=current.time+1;
            if(map[next.x][next.y]=='x')
               next.time++;
            if(check(next.x,next.y)&&map[next.x][next.y]!='#'&&next.time