HDU 3681 Prison Break

HDU_3681

    由于Y和G加起来不到15个，那么可以预先将F、Y、G之间的最短路处理出来，然后化归成TSP问题来做。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAXD 20
int N, M, dis[MAXD][MAXD], g[MAXD][MAXD], mark, Y, G, sx, sy, vis[MAXD][MAXD];
int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
char b[MAXD][MAXD];
int f[1 << 15 | 10][MAXD];
void init()
{
    int i, j;
    Y = 1;
    for(i = 1; i <= N; i ++)
    {
        scanf("%s", b[i] + 1);
        for(j = 1; j <= M; j ++)
        {
            if(b[i][j] == 'F') g[i][j] = 0, sx = i, sy = j;
            else if(b[i][j] == 'S') g[i][j] = -1;
            else if(b[i][j] == 'Y') g[i][j] = Y ++;
            else if(b[i][j] == 'D')    g[i][j] = -2;
        }
    }
    mark = (1 << Y) - 2, G = Y;
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            if(b[i][j] == 'G') g[i][j] = G ++;
}
inline int inside(int x, int y)
{
    return x >= 1 && x <= N && y >= 1 && y <= M;
}
void dfs(int x, int y)
{
    int i, nx, ny;
    vis[x][y] = 1;
    for(i = 0; i < 4; i ++)
    {
        nx = x + dx[i], ny = y + dy[i];
        if(inside(nx, ny) && !vis[nx][ny] && g[nx][ny] != -2) dfs(nx, ny);
    }
}
int check()
{
    int i, j;
    memset(vis, 0, sizeof(vis));
    dfs(sx, sy);
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            if(b[i][j] == 'Y' && !vis[i][j]) return 0;
    return 1;
}
void bfs(int sx, int sy)
{
    int i, j, x, y, nx, ny, id = g[sx][sy];
    memset(vis, -1, sizeof(vis));
    std::queue <int> q;
    vis[sx][sy] = 0;
    q.push(sx * (M + 1) + sy);
    while(!q.empty())
    {
        x = q.front() / (M + 1), y = q.front() % (M + 1), q.pop();
        if(g[x][y] >= 0) dis[id][g[x][y]] = vis[x][y];
        for(i = 0; i < 4; i ++)
        {
            nx = x + dx[i], ny = y + dy[i];
            if(inside(nx, ny) && vis[nx][ny] == -1 && g[nx][ny] != -2)
                vis[nx][ny] = vis[x][y] + 1, q.push(nx * (M + 1) + ny);
        }
    }
}
int dp(int limit)
{
    int i, j, k;
    memset(f, -1, sizeof(f));
    f[1][0] = limit;
    for(i = 1; i < (1 << G); i ++)
        for(j = 0; j < G; j ++)
        {
            if(f[i][j] == -1) continue;
            if((i & mark) == mark) return 1;
            for(k = 0; k < G; k ++)
                if((i & 1 << k) == 0 && f[i][j] >= dis[j][k])
                {
                    if(k >= Y) f[i | 1 << k][k] = limit;
                    else f[i | 1 << k][k] = std::max(f[i | 1 << k][k], f[i][j] - dis[j][k]);
                }
        }
    return 0;
}
void solve()
{
    int i, j;
    if(Y == 1)
    {
        printf("0\n");
        return ;
    }
    if(!check())
    {
        printf("-1\n");
        return ;
    }
    memset(dis, 0x3f, sizeof(dis));
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            if(g[i][j] >= 0) bfs(i, j);
    int min = 0, max = 1000, mid;
    for(;;)
    {
        mid = min + max + 1 >> 1;
        if(mid == max) break;
        if(dp(mid)) max = mid;
        else min = mid;
    }
    printf("%d\n", mid);
}
int main()
{
    while(scanf("%d%d", &N, &M), N || M)
    {
        init();
        solve();
    }
    return 0;
}