hdu 4771, 枚举

Stealing Harry Potter's Precious

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 131    Accepted Submission(s): 68

Problem Description

　 　Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Vernon never allows such magic things in his house. So Harry has to deposit his precious in the Gringotts Wizarding Bank which is owned by some goblins. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)..... A 3×4 bank grid is shown below:



　 　Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers' properties, so they live in the indestructible rooms and put customers' properties in vulnerable rooms. Harry Potter's precious are also put in some vulnerable rooms. Dudely wants to steal Harry's things this holiday. He gets the most advanced drilling machine from his father, uncle Vernon, and drills into the bank. But he can only pass though the vulnerable rooms. He can't access the indestructible rooms. He starts from a certain vulnerable room, and then moves in four directions: north, east, south and west. Dudely knows where Harry's precious are. He wants to collect all Harry's precious by as less steps as possible. Moving from one room to another adjacent room is called a 'step'. Dudely doesn't want to get out of the bank before he collects all Harry's things. Dudely is stupid.He pay you $1,000,000 to figure out at least how many steps he must take to get all Harry's precious.

 

Input

　　There are several test cases.
　　In each test cases:
　　The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 100).
　 　Then a N×M matrix follows. Each element is a letter standing for a room. '#' means a indestructible room, '.' means a vulnerable room, and the only '@' means the vulnerable room from which Dudely starts to move.
　　The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter's precious in the bank.
　 　In next K lines, each line describes the position of a Harry Potter's precious by two integers X and Y, meaning that there is a precious in room (X,Y).
　　The input ends with N = 0 and M = 0

 

Output

　　For each test case, print the minimum number of steps Dudely must take. If Dudely can't get all Harry's things, print -1.

 

Sample Input

2 3 ##@ #.# 1 2 2 4 4 #@## .... #### .... 2 2 1 2 4 0 0

 

Sample Output

-1 5

 

Source

2013 Asia Hangzhou Regional Contest

 

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参考了http://blog.csdn.net/u011329714/article/details/14648829

枚举到k个点的次序，比如有三个点，a1, a2, a3, 源点为a0, 则可以（a0 -> a2） -> (a2 -> a1) -> (a1 -> a3), 也就是2, 1, 3这个排列， 所以求全排列并bfs即可。

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<queue>

using namespace std;

#define LL long long
#define MAXN 111
char g[MAXN][MAXN];
int n, m, k;
int sx, sy;

struct node{
    int stp, x, y;
}target[5];
#define cint const int
cint dx[]={-1, 1, 0, 0}, dy[]={0, 0, -1, 1};

void reset(){           //初始化，相当于memeset(vis, false, ...)
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++) if(g[i][j] == 'v')
            g[i][j]='.';
}

int bfs(int *p){
    queue<node> q;
    reset();
    q.push((node){0, sx, sy});      g[sx][sy]='v';      //visited
    for(int i=0; !q.empty(); ){
        node tnode = q.front();     q.pop();
        for(int j=0; j<4; j++){
            int nx = tnode.x + dx[j], ny = tnode.y + dy[j], stp = tnode.stp + 1;
            if(nx<0 || ny<0 || nx>=n || ny>=m || g[nx][ny]=='#' || g[nx][ny]=='v') continue;
            bool f = nx==target[p[i]].x && ny==target[p[i]].y;      //走到当前想到的地方
            if(f){              //到了，初始化。
                while(!q.empty()) q.pop();
                reset();
                if(++i == k) return stp;
            }
            q.push((node){stp, nx, ny});    g[nx][ny]='v';
            if(f) break;
        }
    }
    return -1;              //目地不可达。
}

int main(){
//    freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
    while(scanf(" %d %d", &n, &m)==2 && n){
        int i, j, p[5], cas=1;
        for(i=0; i<n; i++){
            scanf(" %s", g[i]);
            for(j=0; j<m; j++) if(g[i][j] == '@')
                sx=i, sy=j;
        }
        scanf(" %d", &k);
        for(i=0; i<k; i++){
            scanf(" %d %d", &target[i].x, &target[i].y);
            target[i].x--;       target[i].y--;
            cas = cas * (i + 1);                // k!
            p[i] = i;
        }
        int ans = -1;
        while(cas--){
            int tmp = bfs(p);
            if(tmp>-1 && (tmp < ans || ans==-1)) ans = tmp; //合法的最小值
            next_permutation(p, p+k);
        }
        printf("%d
", ans);
    }
    return 0;
}

 每次贪心可达的最近目的点的WA代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<cmath>
#include<fstream>
#include<iomanip>

using namespace std;

#define LL long long
#define MAXN 111

int n, m, k, tofind;
char g[MAXN][MAXN];

struct node{
    int stp, x, y;
};

#define cint const int
cint dx[]={-1, 1, 0, 0},
     dy[]={0, 0, -1, 1};

bool cango(int x, int y){
    if(x<0 || x>=n || y<0 || y>=m || g[x][y]=='#' || g[x][y]=='v') return false;
    return true;
}

void reset(){
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++) if(g[i][j] == 'v')
            g[i][j] = '.';
}

int bfs(){
    queue<node> q;
    int i, j;
    for(i=0; i<n; i++){
        for(j=0; j<m; j++) if(g[i][j] == '@')
            break;
        if(j<m) break;
    }
    q.push((node){0, i, j});     g[i][j]='v';
    while(!q.empty()){
        node t = q.front();     q.pop();
        for(i=0; i<4; i++){
            int nx = dx[i] + t.x, ny = dy[i] + t.y, stp = t.stp + 1;
            if(cango(nx, ny)){
                bool f = (g[nx][ny] == 'q');
                if(f){
                    if(++tofind == k) return stp;
                    while(!q.empty()) q.pop();
                    reset();
                }
                q.push((node){stp, nx, ny});    g[nx][ny]='v';
                if(f) break;
            }
        }
    }
    return -1;
}

int main(){
//    freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
    while(scanf(" %d %d", &n, &m)==2 && n){
        int i;
        for(i=0; i<n; i++) scanf(" %s", g[i]);
        scanf(" %d", &k);
        for(tofind = i = 0; i<k; i++){
            int x, y;
            scanf(" %d %d", &x, &y);    x--;    y--;
            if(g[x][y] == '@' || g[x][y]=='q') tofind++;
            else g[x][y] = 'q';
        }
        int ans =(tofind==k ? 0 : bfs());
        printf("%d
", ans);
    }
    return 0;
}