【BFS】HDU1026Ignatius and the Princess I

时间：2016-05-12 21:42:34 阅读：205 评论：0 收藏：0 [点我收藏+]

标签：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1026

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

代码：

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m;
char Map[150][150];
int vis[150][150];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
typedef struct NODE
{
    int x,y;
}NODE;
struct node
{
    int s;
    NODE now;
    NODE pre;
    friend bool operator <(node a,node b)
    {
        return a.s>b.s;
    }
}a[150][150];
void init()
{
    memset(vis,0,sizeof(vis));
}
bool check(int x,int y)
{
    return !vis[x][y]&&x>=0&&y>=0&&x<n&&y<m&&(Map[x][y]!='X');
}
void print(node st)
{
    cout<<"It takes "<<st.s<<" seconds to reach the target position, let me show you the way."<<endl;
    NODE route[10050];
    route[0]=st.now;
    int cnt=1;
    while(st.now.x!=0||st.now.y!=0){
        st=a[st.pre.x][st.pre.y];
        route[cnt++]=st.now;
    }
    int t=1;
    for(int i=cnt-1;i>0;i--){
        if(Map[route[i].x][route[i].y]!='.'){
            char tt=Map[route[i].x][route[i].y];
            while(tt!='0'){
                cout<<t++<<"s:FIGHT AT ("<<route[i].x<<','<<route[i].y<<")"<<endl;
                tt--;
            }
        }
        cout<<t++<<"s:("<<route[i].x<<','<<route[i].y<<")->("<<route[i-1].x<<','<<route[i-1].y<<')'<<endl;
    }
    if(Map[route[0].x][route[0].y]!='.'){
        char tt=Map[route[0].x][route[0].y];
        while(tt!='0'){
            cout<<t++<<"s:FIGHT AT ("<<route[0].x<<','<<route[0].y<<")"<<endl;
            tt--;
        }
    }
    return;
}
void bfs()
{
    priority_queue<node>q;
    node st;
    st.s=0;
    st.now.x=0;
    st.now.y=0;
    st.pre.x=0;
    st.pre.y=0;
    a[0][0]=st;
    vis[0][0]=1;
    q.push(st);
    while(!q.empty()){
        st=q.top();
        q.pop();
        if(st.now.x==n-1&&st.now.y==m-1){
            print(st);
            //cout<<st.s<<endl;
            return;
        }
        for(int i=0;i<4;i++){
            node nt=st;
            nt.pre.x=st.now.x;
            nt.pre.y=st.now.y;
            nt.now.x+=dir[i][0];
            nt.now.y+=dir[i][1];
            if(check(nt.now.x,nt.now.y)){
                vis[nt.now.x][nt.now.y]=1;
                a[nt.now.x][nt.now.y]=nt;
                int t=1;
                if(Map[nt.now.x][nt.now.y]!='.'){
                    t+=Map[nt.now.x][nt.now.y]-'0';
                }
                nt.s+=t;
                q.push(nt);
            }
        }
    }
    cout<<"God please help our poor hero."<<endl;
}
int main()
{
    cin.sync_with_stdio(false);
    while(cin>>n>>m){
        init();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>Map[i][j];
            }
        }
        bfs();
        cout<<"FINISH"<<endl;
    }
    return 0;
}

【BFS】HDU1026Ignatius and the Princess I

标签：

原文地址：http://blog.csdn.net/wlxsq/article/details/51353324

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行