洛谷P3070 [USACO13JAN]岛游记Island Travels

P3070 [USACO13JAN]岛游记Island Travels

题目描述

Farmer John has taken the cows to a vacation out on the ocean! The cows are living on N (1 <= N <= 15) islands, which are located on an R x C grid (1 <= R, C <= 50). An island is a maximal connected group of squares on the grid that are marked as 'X', where two 'X's are connected if they share a side. (Thus, two 'X's sharing a corner are not necessarily connected.)

Bessie, however, is arriving late, so she is coming in with FJ by helicopter. Thus, she can first land on any of the islands she chooses. She wants to visit all the cows at least once, so she will travel between islands until she has visited all N of the islands at least once.

FJ's helicopter doesn't have much fuel left, so he doesn't want to use it until the cows decide to go home. Fortunately, some of the squares in the grid are shallow water, which is denoted by 'S'. Bessie can swim through these squares in the four cardinal directions (north, east, south, west) in order to travel between the islands. She can also travel (in the four cardinal directions) between an island and shallow water, and vice versa.

Find the minimum distance Bessie will have to swim in order to visit all of the islands. (The distance Bessie will have to swim is the number of distinct times she is on a square marked 'S'.) After looking at a map of the area, Bessie knows this will be possible.

给你一张r*c的地图，有’S’,’X’,’.’三种地形，所有判定相邻与行走都是四连通的。我们设’X’为陆地，一个’X’连通块为一个岛 屿，’S’为浅水，’.’为深水。刚开始你可以降落在任一一块陆地上，在陆地上可以行走，在浅水里可以游泳。并且陆地和浅水之间可以相互通行。但无论如何 都不能走到深水。你现在要求通过行走和游泳使得你把所有的岛屿都经过一边。Q：你最少要经过几个浅水区？保证有解。

输入输出格式

输入格式：

  • Line 1: Two space-separated integers: R and C.

  • Lines 2..R+1: Line i+1 contains C characters giving row i of the grid. Deep water squares are marked as '.', island squares are marked as 'X', and shallow water squares are marked as 'S'.
输出格式：
  • Line 1: A single integer representing the minimum distance Bessie has to swim to visit all islands.

输入输出样例

输入样例#1：
5 4 
XX.S 
.S.. 
SXSS 
S.SX 
..SX
输出样例#1：
3

说明

There are three islands with shallow water paths connecting some of them.

Bessie can travel from the island in the top left to the one in the middle, swimming 1 unit, and then travel from the middle island to the one in the bottom right, swimming 2 units, for a total of 3 units.

 

【题解】

很撩人的一道题。第一次见隐式图SPFA，不过还算好理解，看看代码可以理解差不多。

很裸的状压DP。

  1 #include <bits/stdc++.h>
  2 const int INF = 0x3f3f3f3f;
  3 const int MAXN = 15 + 1;
  4 const int MAXR = 50 + 5;
  5 const int MAXC = 50 + 5;
  6 const int dx[4] = {0,0,1,-1};
  7 const int dy[4] = {1,-1,0,0};
  8 
  9 inline void read(int &x)
 10 {
 11     x = 0;char ch = getchar();char c = ch;
 12     while(ch > '9' || ch < '0')c = ch, ch = getchar();
 13     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
 14     if(c == '-')x = -x;
 15 }
 16 inline int min(int a, int b){return a > b ? b : a;}
 17 
 18 char g[MAXR][MAXC];int flag[MAXR][MAXC],b[MAXR][MAXC],num[MAXN];
 19 int r,c;
 20 int d[MAXR][MAXC];
 21 
 22 struct Node
 23 {
 24     int x,y;
 25 }block[MAXN][MAXN];
 26 
 27 std::queue <Node> q;int cnt;
 28 inline void bfs(int sx, int sy)
 29 {
 30     q.push(Node{sx,sy});
 31     b[sx][sy] = 1;
 32     flag[sx][sy] = cnt;
 33     ++ num[cnt];
 34     block[cnt][num[cnt]] = Node{sx, sy};
 35     while(!q.empty())
 36     {
 37         Node now = q.front();
 38         q.pop();
 39         for(int i = 0;i < 4;++ i)
 40         {
 41             int xx = now.x + dx[i];
 42             int yy = now.y + dy[i];
 43             if(xx <= r && xx > 0 && yy <= c && yy > 0 && !b[xx][yy] && g[xx][yy] == 'X')
 44             {
 45                 q.push(Node{xx, yy});
 46                 b[xx][yy] = 1;
 47                 flag[xx][yy] = cnt;
 48                 ++ num[cnt];
 49                 block[cnt][num[cnt]] = Node{xx, yy};
 50             }
 51         }
 52     }
 53 }
 54 
 55 int dis[MAXR][MAXC];
 56 void SPFA(int n)
 57 {
 58     memset(b, 0, sizeof(b));
 59     memset(dis, 0x3f, sizeof(dis));
 60     for(int i = 1;i <= num[n];++ i)
 61     {
 62         b[block[n][i].x][block[n][i].y] = 1;
 63         dis[block[n][i].x][block[n][i].y] = 0;
 64         q.push(block[n][i]);
 65     }
 66     Node now;
 67     while(!q.empty())
 68     {
 69         now = q.front();
 70         q.pop();
 71         b[now.x][now.y] = 0;
 72         for(int i = 0;i < 4;++ i)
 73         {
 74             int xx = now.x + dx[i];
 75             int yy = now.y + dy[i];
 76             if(xx <= 0 || xx > r || yy <= 0 || yy > c || g[xx][yy] == '.')continue;
 77             if(g[xx][yy] == 'X')
 78             {
 79                 if(dis[xx][yy] > dis[now.x][now.y])
 80                 {
 81                     dis[xx][yy] = dis[now.x][now.y];
 82                     if(!b[xx][yy])
 83                     {
 84                         b[xx][yy] = true;
 85                         q.push(Node{xx, yy});
 86                     }
 87                 }
 88                 d[flag[xx][yy]][n] = d[n][flag[xx][yy]] = min(d[n][flag[xx][yy]], min(d[flag[xx][yy]][n],dis[xx][yy]));
 89             }
 90             else if(g[xx][yy] == 'S')
 91             {
 92                 if(dis[xx][yy] > dis[now.x][now.y] + 1)
 93                 {
 94                     dis[xx][yy] = dis[now.x][now.y] + 1;
 95                     if(!b[xx][yy])
 96                     {
 97                         b[xx][yy] = true;
 98                         q.push(Node{xx, yy});
 99                     }
100                 }
101             }
102         }
103     }
104 }
105 
106 inline void init()
107 {
108     read(r);read(c);
109     for(int i = 1;i <= r;++ i)
110         scanf("%s", g[i] + 1);
111     for(int i = 1;i <= r;++ i)
112         for(int j = 1;j <= c;++ j)
113             if(!b[i][j] && g[i][j] == 'X')
114             {
115                 ++ cnt; 
116                 bfs(i, j);
117             } 
118     memset(d, 0x3f, sizeof(d));
119     for(int i = 1;i <= cnt;++ i)
120         d[i][i] = 0, SPFA(i);
121 }
122 
123 //状压方程：F[S][j]表示走过点集S到达j的最短路径
124 //转移：F[S/j][k] + dis[k,j] 
125 int dp[(1 << MAXN)][MAXN],ans;
126 
127 inline void DP()
128 {
129     memset(dp, 0x3f, sizeof(dp));
130     int tmp = (1 << cnt);
131     for(int i = 1;i <= cnt;++ i)
132         dp[1 << (i - 1)][i] = 0;
133     for(int S = 1;S < tmp;++ S)
134     {
135         for(int i = 1;i <= cnt;++ i)
136         {
137             if(!(S & (1 << (i - 1))))continue;
138             for(int j = 1;j <= cnt;++ j)
139             {
140                 if(j == i || !(S & (1 << (j - 1))))continue;
141                 dp[S][i] = min(dp[S][i], dp[S ^ (1 << (i - 1))][j] + d[j][i]);
142             }
143         }
144     }
145 }
146 
147 inline void put()
148 {
149     int S = (1 << cnt) - 1;
150     ans = INF;
151     for(int j = 1;j <= cnt;++ j)
152         ans = min(ans, dp[S][j]);
153     printf("%d", ans);
154 } 
155 
156 int main()
157 {
158     init(); 
159     DP();
160     put();
161     return 0;
162 }
View Code

 

 

posted @ 2017-07-06 15:10  嘒彼小星  阅读(283)  评论(0)  编辑  收藏  举报