HDU 1242 Rescue（求最短時間救出同伴，BFS+DP）

Rescue

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

 

題目大意：

你的朋友被抓進監獄，現在你要找到你的朋友，在途中你會遇到牆（’#‘），不能走，路（‘.’）可以走，花時爲1秒，士兵（‘X’）,花時間爲2秒，求最短時間。

解題思路:

BFS和DP，不能單純的只用BFS，因爲有的一步是要花時間2，有的爲1，不能確保當前的就是最優的，所有有要用到DP。

代碼;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

struct node{
    int i,j;
    node(int i0=0,int j0=0){
        i=i0,j=j0;
    }
};

const int dirJ[4]={0,1,0,-1};
const int dirI[4]={1,0,-1,0},maxN=220;
int n,m,is,js,ie,je,time[maxN][maxN];
char str[maxN][maxN];

void bfs(){
    queue <node> path;
    time[is][js]=0;
    path.push(node(is,js));
    while(!path.empty()){
        node s=path.front();
        path.pop();
        for(int k=0;k<4;k++){
            int di=s.i+dirI[k],dj=s.j+dirJ[k];
            if(di>=n||dj>=m||di<0||dj<0) continue;
            if(str[di][dj]=='#') continue;
            int tmp=time[s.i][s.j]+1;
            if(str[di][dj]=='x') tmp++;
            if( tmp<time[di][dj] || time[di][dj]==-1 ){//帶點dp優化.
                path.push(node(di,dj));
                time[di][dj]=tmp;
            }
        }
    }
    if(time[ie][je]==-1) printf("Poor ANGEL has to stay in the prison all his life.\n");
    else printf("%d\n",time[ie][je]);
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(time,-1,sizeof(time));
        for(int i=0;i<n;i++){
            scanf("%s",str[i]);//建議這樣。
            for(int j=0;j<m;j++){
                if(str[i][j]=='r'){is=i,js=j;}
                if(str[i][j]=='a'){ie=i,je=j;}
            }
        }
        bfs();
    }
    return 0;
}