Ignatius and the Princess I

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

Author

Ignatius.L

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题目大意：

给定1张图，走到“.”需要1步，走到数字除了需要1步，还要停留数字上那么多步，“#”不能走，问你从左上角到右下至少走多少步，并输出路径

解题思路：

简单的BFS，再加上记录前1步可以从终点往前来获得路径。

解题代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

struct node{
    int x,y,step;
    node(int x0=0,int y0=0,int step0=0){
        x=x0;y=y0;step=step0;
    }
};

const int offx[]={1,-1,0,0};
const int offy[]={0,0,-1,1};
int n,m,step[110][110],pre[110][110];
char a[110][110];

void solve(){
    memset(step,-1,sizeof(step));
    memset(pre,-1,sizeof(pre));
    queue <node> q;
    q.push(node(0,0,0));
    step[0][0]=0;
    while(!q.empty()){
        node s=q.front();
        q.pop();
        for(int i=0;i<4;i++){
            int dx=s.x+offx[i],dy=s.y+offy[i],newstep;
            if(dx<0 || dx>=n || dy<0 || dy>=m) continue;
            if(a[dx][dy]=='X') continue;
            if(a[dx][dy]=='.') newstep=s.step+1;
            else newstep=s.step+1+a[dx][dy]-'0';
            if(newstep<step[dx][dy] || step[dx][dy]==-1){
                step[dx][dy]=newstep;
                pre[dx][dy]=i;
                q.push(node(dx,dy,step[dx][dy]));
            }
        }
    }
}

void dfs(int dx,int dy){
    if(pre[dx][dy]==-1) return;
    int sx=dx-offx[pre[dx][dy]],sy=dy-offy[pre[dx][dy]];
    dfs(sx,sy);
    printf("%ds:(%d,%d)->(%d,%d)\n",step[sx][sy]+1,sx,sy,dx,dy);
    if(a[dx][dy]!='.'){
        for(int i=1;i<=a[dx][dy]-'0';i++){
            printf("%ds:FIGHT AT (%d,%d)\n",step[sx][sy]+1+i,dx,dy);
        }
    }
}

void output(){
    if(step[n-1][m-1]==-1){
        printf("God please help our poor hero.\n");
    }else{
        printf("It takes %d seconds to reach the target position, let me show you the way.\n",step[n-1][m-1]);
        dfs(n-1,m-1);
    }
    printf("FINISH\n");
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=0;i<n;i++) scanf("%s",a[i]);
        solve();
        output();
    }
    return 0;
}

HDU 1026 Ignatius and the Princess I （基本算法-BFS）,布布扣,bubuko.com