hdu 1026 Ignatius and the Princess I （BFS+優先隊列）

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Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11700    Accepted Submission(s): 3653
Special Judge

Problem DescriptionThe Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166‘s castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166‘s room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position. 

 

InputThe input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input. 

 

OutputFor each test case, you should output "God please help our poor hero." if Ignatius can‘t reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output. 

 

Sample Input5 6.XX.1...X.2.2...X....XX.XXXXX.5 6.XX.1...X.2.2...X....XX.XXXXX15 6.XX.....XX1.2...X....XX.XXXXX. 

 

Sample OutputIt takes 13 seconds to reach the target position, let me show you the way.1s:(0,0)->(1,0)2s:(1,0)->(1,1)3s:(1,1)->(2,1)4s:(2,1)->(2,2)5s:(2,2)->(2,3)6s:(2,3)->(1,3)7s:(1,3)->(1,4)8s:FIGHT AT (1,4)9s:FIGHT AT (1,4)10s:(1,4)->(1,5)11s:(1,5)->(2,5)12s:(2,5)->(3,5)13s:(3,5)->(4,5)FINISHIt takes 14 seconds to reach the target position, let me show you the way.1s:(0,0)->(1,0)2s:(1,0)->(1,1)3s:(1,1)->(2,1)4s:(2,1)->(2,2)5s:(2,2)->(2,3)6s:(2,3)->(1,3)7s:(1,3)->(1,4)8s:FIGHT AT (1,4)9s:FIGHT AT (1,4)10s:(1,4)->(1,5)11s:(1,5)->(2,5)12s:(2,5)->(3,5)13s:(3,5)->(4,5)14s:FIGHT AT (4,5)FINISHGod please help our poor hero.FINISH 

 

 

 

          題意:就是讓你找到從（0,0）到（n-1,m-1）的最短路徑，並把這條路徑給輸出來，若沒有則輸出God please help our poor hero，而且輸出路徑時如果遇到怪物，與它戰鬥了幾秒也要輸出來。

 

         解題思路：用BFS+優先隊列的方法，我們不斷尋找下一個位置，並把耗時最少的又作為起點，尋找下一個位置，如果能找到，則第一次找到的一定是最短耗時的，但這一題有一個痛點，就是難在如何儲存我找到的最短路徑。我們開一個結構體數組，當我們找到下一個位置的時候，在下一個節點中儲存上一個節點的位置，這樣我們就可以從終點遞推來找到整條路，然後把它儲存在另一個數組中，這樣我們就直接輸出這個數組就可以了。

 

貼出代碼：

 

#include <stdio.h>#include <iostream>#include <queue>using namespace std;int n, m, mark, minx;int visited[105][105];int dir[4][2] = {0, -1, 1, 0, 0, 1, -1, 0};int Footx[10005], Footy[10005];    //作為轉換的路徑數組char map[105][105];struct foot    //儲存路徑的結構體{    int x, y;}Foot[105][105];struct node         //優先隊列結構體，耗時少的優先{    int x, y;    int time;    friend bool operator < (node a, node b)    {        return a.time > b.time;    }};int Judge(int x, int y)     //判斷這個點是否符合{    if(x<0 || x>=n || y<0 || y>=m)        return 1;    if(map[x][y] == ‘X‘ || visited[x][y])        return 1;    return 0;}void BFS(int startx, int starty, int time)     //BFS搜尋最短路{    priority_queue <node> Q;    struct node Node, temp;    Node.x = startx;      Node.y = starty;    Node.time = 0;    visited[startx][starty] = 1;    Q.push(Node);    while(!Q.empty())    {        Node = Q.top();        Q.pop();        if(Node.x == n-1 && Node.y == m-1)        {            minx = Node.time;            mark = 1;            return;        }        for(int i = 0; i<4; i++)        {            temp.x = Node.x+dir[i][0];            temp.y = Node.y+dir[i][1];            temp.time = Node.time+1;            if(Judge(temp.x, temp.y))                continue;            if(map[temp.x][temp.y] == ‘.‘)            {                Foot[temp.x][temp.y].x = Node.x;     //將上一個節點儲存下來                Foot[temp.x][temp.y].y = Node.y;                visited[temp.x][temp.y] = 1;                Q.push(temp);            }            if(map[temp.x][temp.y]>=‘1‘ && map[temp.x][temp.y]<=‘9‘)            {                Foot[temp.x][temp.y].x = Node.x;                Foot[temp.x][temp.y].y = Node.y;                temp.time = temp.time+map[temp.x][temp.y]-‘1‘+1;                visited[temp.x][temp.y] = 1;                Q.push(temp);            }        }    }}void Show(){    printf("It takes %d seconds to reach the target position, let me show you the way.\n", minx);    int k = 0, i =minx, num;    Footx[k] = n-1;    Footy[k] = m-1;    while(i--)       //將儲存下來的路徑轉換出來，我們就可以知道走過了那些點    {        if(Footx[k] == 0 && Footy[k] == 0)            break;        k++;        Footx[k] = Foot[ Footx[k-1] ][ Footy[k-1] ].x;        Footy[k] = Foot[ Footx[k-1] ][ Footy[k-1] ].y;    }    num = 1;    for(int j = k-1; j>=0; j--)    //根據走過的這些點來進行輸出    {        if(map[ Footx[j] ][ Footy[j] ] == ‘.‘)            printf("%ds:(%d,%d)->(%d,%d)\n", num++, Footx[j+1], Footy[j+1], Footx[j], Footy[j]);        if(map[ Footx[j] ][ Footy[j] ]>=‘1‘ && map[ Footx[j] ][ Footy[j] ]<=‘9‘)        {            printf("%ds:(%d,%d)->(%d,%d)\n", num++, Footx[j+1], Footy[j+1], Footx[j], Footy[j]);            for(int t = 1; t<=map[ Footx[j] ][ Footy[j] ]-‘0‘; t++)            {                printf("%ds:FIGHT AT (%d,%d)\n", num++, Footx[j], Footy[j]);            }        }    }}int main(){    while(scanf("%d%d", &n, &m)!=EOF)    {        for(int i = 0; i<n; i++)        {            for(int j = 0; j<m; j++)            {                cin>>map[i][j];                visited[i][j] = 0;            }        }        mark = 0;        BFS(0, 0, 0);        if(mark)            Show();        else            printf("God please help our poor hero.\n");        printf("FINISH\n");    }    return 0;}