poj 3328 Cliff Climbing 题解

做了这道题目，感觉纯搜索更上一次楼啦。

题意：

计算从起点 S 走到 终点 T 的最少耗时，范围 60*30。

走路分双脚，每次动一个脚，每次走一步有 9 个选择，耗时为 graph[x][y]，即 (x, y) 坐标上的数字。其中 ‘X’ 代表不能走，只要有一个脚走到 ‘T’，就表示到终点。终点不止一个！

PS：

起点为 ‘S’，但分左右脚，即起点踏左脚，右脚就在图外，也就是说先走右脚，反之，同理。不止一个起点！

第一次用 queue 直接交了，结果 TLE，改换 priority_queue，AC。

zstu_wangrui 3328Accepted 13860K1250MS C++3601B 2013-07-20 12:38:08

算法关键：dist[60][30][60][30]

/*poj 3328 Cliff Climbing走格子——更上一层楼呀多起点，多终点，分左右，9个方向的走法，格子范围：60 * 30zstu_wangrui3328Accepted13860K1204MSC++2919B2013-07-20 12:57:01*/#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int inf = 100000000;int dd( int x ){ if( x < 0 )return -x;return x; }int dir[9][2] = {{-2, 1}, {-1, 1}, {0, 1}, {1, 1}, {2, 1}, {-1, 2}, {0, 2}, {1, 2}, {0, 3}};struct pos{int x, y;};struct Feet{pos left, right;int cost;bool operator<( Feet e )const{return e.cost < cost;}};int graph[61][31], dist[61][31][61][31];int os, w, h;priority_queue<Feet>q;void init(){int i, j, k, t, x, y;for( i = 1; i <= h; i++ )for( j = 1; j <= w; j++ )for( k = 1; k <= h; k++ )for( t = 1; t <= w; t++ )dist[i][j][k][t] = inf;while( !q.empty() )q.pop();Feet id;for( i = 1; i <= h; i++ ){for( j = 1; j <= w; j++ ){if( graph[i][j] == -1 ){id.left.x = i;id.left.y = j;id.right.x = id.right.y = -1;id.cost = 0;q.push( id );id.left.x = id.left.y = -1;id.right.x = i;id.right.y = j;id.cost = 0;q.push( id );}}}}int bfs(){init();Feet now, to;int i, x, y, xx, yy;while( !q.empty() ){now = q.top();q.pop();if( now.left.x != -1 && now.right.x != -1 && (graph[now.left.x][now.left.y] == 0 || graph[now.right.x][now.right.y] == 0) )return now.cost;x = now.left.x;y = now.left.y;if( x != -1 && y != -1 )for( i = 0; i < 9; i++ ){xx = x + dir[i][0];yy = y + dir[i][1];if( xx >= 1 && xx <= h && yy >= 1 && yy <= w && graph[xx][yy] != -2 && graph[xx][yy] != -1 && dist[x][y][xx][yy] > now.cost + graph[xx][yy] ){dist[x][y][xx][yy] = now.cost + graph[xx][yy];to.cost = now.cost + graph[xx][yy];to.left = now.left;to.right.x = xx;to.right.y = yy;q.push( to );}}x = now.right.x;y = now.right.y;if( x != -1 && y != -1 )for( i = 0; i < 9; i++ ){xx = x + dir[i][0];yy = y - dir[i][1];if( xx >= 1 && xx <= h && yy >= 1 && yy <= w && graph[xx][yy] != -2 && graph[xx][yy] != -1 && dist[xx][yy][x][y] > now.cost + graph[xx][yy] ){dist[xx][yy][x][y]  = now.cost + graph[xx][yy];to.cost = now.cost + graph[xx][yy];to.left.x = xx;to.left.y = yy;to.right = now.right;q.push( to );}}}return -1;}int main(){int i, j;while( ~scanf( "%d%d", &w, &h ), w + h ){char sign[3];for( i = 1; i <= h; i++ ){for( j = 1; j <= w; j++ ){scanf( "%s", sign );if( sign[0] == 'X' )graph[i][j] = -2;else if( sign[0] == 'T' )graph[i][j] = 0;else if( sign[0] == 'S' )graph[i][j] = -1;elsegraph[i][j] = sign[0] - '0';}}printf( "%d\n", bfs() );}return 0;}