Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12750    Accepted Submission(s): 4034

Special Judge

Problem Description

The 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.

Input

The 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.

Output

For 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 Input

5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.

Sample Output

It 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)
FINISH
It 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)
FINISH
God please help our poor hero.
FINISH

题意：给你一个图，里面由‘.’，‘X’和数字组成，从（0， 0）出发，问能不能到达（n-1， m-1）点。如果能到达， 每走一步，都要花1秒的时间，问最短的时间，并将路径输出来。

这道题题意不难理解，难的是一定要找到最短的那个路径，用一般的BFS肯定wA,不要问我为什么，说多了都是泪~~~

（一般的BFS虽然说能判断能否达到，但不一定是最短距离。）

分析：我们建造一个二维数组，用来记录路径以及从（0， 0）点到达该点的最短距离，每次搜索到一个满足条件的点，都让到该点的时间与原来到达该点的时间相比较，如果是当前的路径小于原来到该点的时间，就更新时间和路径，同时入队列。。最后就是每个点都是最优解了

心得：好题！！！让窝了解了一种如何找最优路径的方法

话说起来，这也是我AC的第一道special judge！！！

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <queue>
const int M = 105;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1 ,0, 0};
using namespace std;

struct node{
    int x, y, prex, prey;  //prex，prey是上一个点的坐标
    int cost;
}s[M][M];
char map[M][M];
int n, m;

void init(){
    int i,j;
    for(i = 0; i < M; i ++){
        for(j = 0; j < M; j ++)
            s[i][j].cost = -1;
    }
}

int limit(int x, int y){
    return (x>=0&&x<n&&y>=0&&y<m&&map[x][y] != 'X');
}

void out(){   //最后输出的时候用栈来保存路径
    printf("It takes %d seconds to reach the target position, let me show you the way.\n", s[n-1][m-1].cost);
    stack<node > S;
    node a = s[n-1][m-1];
    S.push(a);
    while(1){
        if(a.x == 0&&a.y == 0) break;
        a = s[a.prex][a.prey];
        S.push(a);
    }
    a = S.top(); S.pop();
    int t = 1;
    while(!S.empty()){
        node b = S.top(); S.pop();
        printf("%ds:(%d,%d)->(%d,%d)\n", t++, a.x, a.y, b.x, b.y);
        int temp = b.cost-a.cost-1;
        while(temp--){
            printf("%ds:FIGHT AT (%d,%d)\n", t++, b.x, b.y);
        }
        a = b;
    }
}

void bfs(){
    int i;
    node st;
    st.x = st.y = st.prex = st.prey = st.cost = 0;
    s[0][0] = st;
    queue<node >q;
    q.push(st);
    while(!q.empty()){
        node temp = q.front();
        q.pop();
        for(i = 0; i < 4; i ++){
            node cur = temp;
            cur.x += dx[i];
            cur.y += dy[i];
            if(!limit(cur.x, cur.y)) continue;
            if(map[cur.x][cur.y] == '.') ++cur.cost;
            else cur.cost += (map[cur.x][cur.y]-'0'+1);
            if(cur.cost < s[cur.x][cur.y].cost||s[cur.x][cur.y].cost == -1){
                s[cur.x][cur.y] = cur;
                s[cur.x][cur.y].prex = temp.x; s[cur.x][cur.y].prey = temp.y;
                q.push(cur);
            }
        }
    }
    if(s[n-1][m-1].cost == -1){
        printf("God please help our poor hero.\n"); return;
    }
    else out();
}

int main(){
    while(scanf("%d%d", &n, &m) == 2){
        init();
        for(int i = 0; i < n; i ++){
            scanf("%s", map[i]);
        }
        bfs();
        printf("FINISH\n");
    }
    return 0;
}