hdu3681_状压dp+bfs+二分答案

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3681

 题意：n*m的矩阵，一个人带着一块满电池从 'F' 出发，走遍所有的 'Y'(开关) ,可以从四个方向（上下左右）， 每走一步消耗一格电，'S'是空地(可走)，'G'是加油站(把电池加满后变为空地)，'D'是不能走的禁地。问关掉所有的开关最少最少带的电池是多少格的？ 如果不存在输出 '-1'

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

const int N = 16, M = 1 << N;
int dp[M][N], k, n, m, first, k2;
int dis[N][N], dist[N][N];//dis[x][y]为某个特殊点到(x, y)的距离, dist[i][j]为(num[i].x, num[i].y)到(num[j].x, num[j].y)的距离
char a[20][20];
struct node {
    int x, y;
}num[20];
int to[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
void bfs(int x, int y)
{
    queue <int> q1, q2;
    q1.push(x), q2.push(y);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            dis[i][j] = INF;
    dis[x][y] = 0;
    while(q1.size())
    {
        int x1 = q1.front(), y1 = q2.front();
        q1.pop();
        q2.pop();
        for(int i = 0; i < 4; i++)
        {
            int x2 = x1 + to[i][0], y2 = y1 + to[i][1];
            if(x2 < 0 || x2 >= n || y2 < 0 || y2 >= m || a[x2][y2] == 'D')
                continue;
            if(dis[x2][y2] > dis[x1][y1] + 1)
            {
                dis[x2][y2] = dis[x1][y1] + 1;
                q1.push(x2), q2.push(y2);
            }
        }
    }
}
void init()
{    
    for(int i = 0; i < k2; i++)
    {
        bfs(num[i].x, num[i].y);//bfs点(num[i].x, num[i].y)到各个点之间的距离
        for(int j = 0; j < k2; j++)
        {
            if(i != j){
                dist[i][j] = dis[num[j].x][num[j].y];
                //cout << i << " " << j <<  " "<< dist[i][j] << endl;
            }
        }
    }
}
bool rec(int p)
{
    int res = (1 << k) - 1;
    for(int i = 0; i < (1 << k2); i++)
        for(int j = 0; j < k2; j++)
            dp[i][j] = -1 * INF;
    dp[1 << first][first] = p;
    for(int i = (1 << first); i < (1 << k2); i++)
    {
        for(int j = 0; j < k2; j++)
        {
            if(dp[i][j] < 0 || !(i & (1 << j)))
                continue;
            if((i & res) == res)
                return true;
            for(int l1 = 0; l1 < k2; l1++)
            {
                if(i & (1 << l1))
                    continue;
                int next = i | (1 << l1);
                dp[next][l1] = max(dp[next][l1], dp[i][j] - dist[j][l1]);
                if(dp[next][l1] >= 0 && l1 >= k)
                    dp[next][l1] = p;
            }
        }
    }
    return false;
}
int main()
{
    while(~scanf("%d %d", &n, &m) && (n + m))
    {
        k = 0;        
        for(int i = 0; i < n; i++){
            getchar();
            for(int j = 0; j < m; j++)
            {
                scanf("%c", &a[i][j]);
                if(a[i][j] == 'F' || a[i][j] == 'Y')
                {                    
                    num[k].x = i, num[k].y = j;
                    k++;
                }
                if(a[i][j] == 'F')
                    first = k - 1;
            }
        }
        k2 = k;   
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++)
            {
                if(a[i][j] == 'G')
                {                    
                    num[k2].x = i, num[k2].y = j;
                    k2++;
                }
            }
        }          
        init(); 
        int l = 0, r = 300, flag = 0;
        while(l <= r)
        {
            int mid = (l + r) / 2;
            if(rec(mid))
            {
                flag = 1;
                r = mid - 1;
            }
            else
                l = mid + 1;
            //cout << l << " " << r << endl;
        }        
        if(!flag)
            printf("-1
");
        else
            printf("%d
", l);        
    }
    return 0;
}