HDU4856 Tunnels —— BFS + 状压DP

最新推荐文章于 2019-08-17 12:14:34 发布

原创最新推荐文章于 2019-08-17 12:14:34 发布 · 454 阅读

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DP——状态压缩

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本文提供了一道来自杭电OJ编号为4856的题目解决方案，采用图搜索与动态规划相结合的方法求解最短路径问题。通过BFS遍历地图并记录各隧道间距离，利用DP算法寻找从任意起点到所有终点的最小代价。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4856

题解：……

类似的DP：http://blog.youkuaiyun.com/dolfamingo/article/details/71817743

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define ms(a, b)  memset((a), (b), sizeof(a))
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 20;

struct tunnel
{
    int x1, y1, x2, y2;
}t[maxn];

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

int n,m;
char maze[maxn][maxn];
int dis[maxn][maxn], dp[1<<15][maxn], vis[maxn][maxn];
int d[4][2] = {1,0,-1,0,0,1,0,-1};

int bfs(int st, int en)
{
    queue<node>q;
    while(!q.empty()) q.pop();
    ms(vis,0);
    vis[t[st].x2][t[st].y2] = 1;

    node now, e;
    now.x = t[st].x2; now.y = t[st].y2; now.dis = 0;
    q.push(now);

    while(!q.empty())
    {
        now = q.front();
        q.pop();

        if(now.x==t[en].x1 && now.y==t[en].y1)
            return now.dis;

        for(int i = 0; i<4; i++)
        {
            int x = now.x + d[i][0];
            int y = now.y + d[i][1];
            int dist = now.dis + 1;

            if(x<1 || x>n || y<1 || y>n || maze[x][y]=='#' || vis[x][y]) continue;

            vis[x][y] = 1;
            e.x = x; e.y = y; e.dis = dist;
            q.push(e);
        }
    }
    return INF;
}

void solve()
{
    for(int s = 0; s<(1<<m); s++)
    for(int j = 0; j<m; j++)
        dp[s][j] = INF;

    for(int i = 0; i<m; i++)
        dp[1<<i][i] = 0;

    for(int s = 1; s<(1<<m); s++)
    {
        for(int j = 0; j<m; j++)
        {
            if(s&(1<<j) && dp[s][j]!=INF)
            for(int k = 0; k<m; k++)
            {
                if( !(s&(1<<k)) && dis[k][j]!=INF)
                {
                    int tmp = s|(1<<k);
                    dp[tmp][k] = min(dp[tmp][k],dp[s][j] + dis[k][j]);
                }
            }
        }
    }

    int ans = INF;
    for(int i = 0; i<m; i++)
        ans = min(ans,dp[(1<<m)-1][i]);

    if(ans!=INF)
        printf("%d\n",ans);
    else
        puts("-1");
}

void build()
{
    for(int i = 0; i<m; i++)
    for(int j = 0; j<m; j++)
    {
        if(i==j)
            dis[i][j] = 0;
        else
            dis[i][j] = bfs(i,j);
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i = 1; i<=n; i++)
            scanf("%s",maze[i]+1);

        for(int i = 0; i<m; i++)
            scanf("%d%d%d%d",&t[i].x1, &t[i].y1, &t[i].x2, &t[i].y2);

        build();
        solve();
    }
}