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                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

13  

 

題意：X代表衛兵，a代表終點，r代表起始點，.代表路，#代表牆

路花費一秒，x花費兩秒

問到達終點的最少時間

思路：BFS+優先佇列的果題

 

#include <stdio.h>#include <string.h>#include <queue>using namespace std;struct node{    int x,y,step;    friend bool operator<(node n1,node n2)    {        return n2.step<n1.step;    }};int n,m,vis[205][205];char map[205][205];int x1,x2,y1,y2;int to[4][2] = {1,0,-1,0,0,1,0,-1};int check(int x,int y){    if(x<0 || y<0 || x>=n || y>=m || !vis[x][y] || map[x][y] == '#')        return 1;    return 0;}int bfs(){    int i;    priority_queue<node> Q;    node a,next;    a.x = x1;    a.y = y1;    a.step = 0;    Q.push(a);    vis[x1][y1] = 0;    while(!Q.empty())    {        a = Q.top();        Q.pop();        if(a.x == x2 && a.y == y2)            return a.step;        for(i = 0; i<4; i++)        {            next = a;            next.x+=to[i][0];            next.y+=to[i][1];            if(check(next.x,next.y))//判斷                continue;            next.step++;            if(map[next.x][next.y] == 'x')//衛兵處多花費了一秒                next.step++;            if(vis[next.x][next.y]>=next.step)//存入最小時間            {                vis[next.x][next.y] = next.step;                Q.push(next);            }        }    }    return 0;}int main(){    int i,j;    while(~scanf("%d%d",&n,&m))    {        for(i = 0; i<n; i++)        {            scanf("%s",map[i]);            for(j = 0; map[i][j]; j++)            {                if(map[i][j] == 'r')                {                    x1 = i;                    y1 = j;                }                else if(map[i][j] == 'a')                {                    x2 = i;                    y2 = j;                }            }        }        memset(vis,1,sizeof(vis));        int ans = 0;        ans = bfs();        if(ans)            printf("%d\n",ans);        else            printf("Poor ANGEL has to stay in the prison all his life.\n");    }    return 0;}

 

           

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