ZOJ-3103 Cliff Climbing

Cliff Climbing

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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At 17:00, special agent Jack starts to escape from the enemy camp. There is a cliff in between the camp and the nearest safety zone. Jack has to climb the almost vertical cliff by stepping his feet on the blocks that cover the cliff. The cliff has slippery blocks where Jack has to spend time to take each step. He also has to bypass some blocks that are too loose to support his weight. Your mission is to write a program that calculates the minimum time to complete climbing.

Figure D-1 shows an example of cliff data that you will receive. The cliff is covered with square blocks. Jack starts cliff climbing from the ground under the cliff, by stepping his left or right foot on one of the blocks marked with 'S' at the bottom row. The numbers on the blocks are the "slippery levels". It takes t time units for him to safely put his foot on a block marked with t, where 1 <= t <= 9. He cannot put his feet on blocks marked with 'X'. He completes the climbing when he puts either of his feet on one of the blocks marked with 'T' at the top row.

Figure D-1: Example of Cliff Data

Jack's movement must meet the following constraints. After putting his left (or right) foot on a block, he can only move his right (or left, respectively) foot. His left foot position (lx, ly) and his right foot position (rx, ry) should satisfy lx < rx and | lx - rx | + | ly - ry | <= 3. This implies that, given a position of his left foot in Figure D-2 (a), he has to place his right foot on one of the nine blocks marked with blue color. Similarly, given a position of his right foot in Figure D-2 (b), he has to place his left foot on one of the nine blocks marked with blue color.

Figure D-2: Possible Placements of Feet

Input

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows:

w h
s(1,1) ... s(1,w)
s(2,1) ... s(2,w)
... 
s(h,1) ... s(h,w)

The integers w and h are the width and the height of the matrix data of the cliff. You may assume 2 <= w <= 30 and 5 <= h <= 60. Each of the following h lines consists of w characters delimited by a space. The character s(y, x) represents the state of the block at position (x, y) as follows:

• 'S': Jack can start cliff climbing from this block.
• 'T': Jack finishes climbing when he reaches this block.
• 'X': Jack cannot put his feet on this block.
• '1' - '9' (= t): Jack has to spend t time units to put either of his feet on this block.

You can assume that it takes no time to put a foot on a block marked with 'S' or 'T'.

Output

For each dataset, print a line only having a decimal integer indicating the minimum time required for the cliff climbing, when Jack can complete it. Otherwise, print a line only having "-1" for the dataset. Each line should not have any characters other than these numbers.

Sample Input

6 64 4 X X T T4 7 8 2 X 73 X X X 1 81 2 X X X 61 1 2 4 4 7S S 2 3 X X2 10T 11 X1 X1 X1 11 X1 X1 11 XS S2 10T X1 X1 X1 X1 11 X1 X1 11 XS S10 10T T T T T T T T T TX 2 X X X X X 3 4 X9 8 9 X X X 2 9 X 97 7 X 7 3 X X 8 9 X8 9 9 9 6 3 X 5 X 58 9 9 9 6 X X 5 X 58 6 5 4 6 8 X 5 X 58 9 3 9 6 8 X 5 X 58 3 9 9 6 X X X 5 XS S S S S S S S S S10 72 3 2 3 2 3 2 3 T T1 2 3 2 3 2 3 2 3 23 2 3 2 3 2 3 2 3 43 2 3 2 3 2 3 2 3 53 2 3 1 3 2 3 2 3 52 2 3 2 4 2 3 2 3 5S S 2 3 2 1 2 3 2 30 0

Sample Output

125-12212

————————————————————中秋后的分割线————————————————————

前言：这其实是中秋节之前做的一道题。想把中秋节当做分界线，今后要脱胎换骨！

思路：不必想太多，这就是一道SPFA的题目。在建图的时候，邻接矩阵存图，只要不是X，而且像是走迷宫一样，另一点可以直接到达，而且没有超过迷宫的边界的话，就存在一条边。但是因为左右脚是交替的，所以需要两张图，一张左脚图，另一张右脚图，松弛操作变成：

if(dis[左][v] > dis[右][u] + mat.val)

        dis[左][v] = dis[右][u] + mat.val;

也就是说， 每次用一只脚松弛另一只脚，交替进行即可。

因为存在多个起点，所以所有起点全部入队。因为存在多个终点，所以取左右脚到达每个终点的最小值。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define INF 0x3f3f3f3fusing namespace std;/****************************************/#define LIM nx >= 0 && nx < h && ny >= 0 && ny < wconst int dir[2][9][2] = {{-2, -1, -1, -1, -1, -2, 0, -1, 0, -2, 0, -3, 1, -1, 1, -2, 2, -1},   {-2, 1, -1, 1, -1, 2, 0, 1, 0, 2, 0, 3, 1, 1, 1, 2, 2, 1}};const int N = 65*35;int w, h, dis[2][N], q[N], ans[N];bool inq[2][N];char mat[65][35];void spfa(int rear){int fron = 0;while(fron < rear) {int u = q[fron%N]; fron++;int x = u / w, y = u % w;for(int f = 0; f < 2; f++) {if(inq[1-f][u]) {//右脚在队列当中inq[1-f][u] = false;for(int d = 0; d < 9; d++) {int nx = x + dir[f][d][0], ny = y + dir[f][d][1];//踏出左脚if(LIM && mat[nx][ny] != 'X') {int v = nx*w + ny, val;if(mat[nx][ny] == 'S' || mat[nx][ny] == 'T') val = 0;else val = mat[nx][ny] - '0';if(dis[f][v] > dis[1-f][u] + val) {dis[f][v] = dis[1-f][u] + val;if(!inq[f][v]) {//新的左脚inq[f][v] = true;q[rear%N] = v; rear++;}}}}}}}}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifwhile(scanf("%d%d", &w, &h), w||h) {int rear = 0;for(int i = 0; i <= h*w; i++) {dis[0][i] = dis[1][i] = INF;inq[0][i] = inq[1][i] = 0;}int cnt = 0;for(int i = 0 ; i < h; i++) {for(int j = 0; j < w; j++) {scanf(" %c", &mat[i][j]);int u = i*w + j;if(mat[i][j] == 'S') {q[rear%N] = u; rear++;dis[0][u] = dis[1][u] = 0;inq[0][u] = inq[1][u] = true;}//起点全部入队if(mat[i][j] == 'T') {ans[cnt++] = i*w + j;}}}spfa(rear);int ret = INF;for(int i = 0; i < cnt; i++) {ret = min(ret, min(dis[0][ans[i]], dis[1][ans[i]]));}//取所有终点的左脚、右脚到达的最小值if(ret == INF) puts("-1");else printf("%d\n", ret);}return 0;}