nyoj 58-最少步數
阿新 • • 發佈:2019-02-13
最少步數
時間限制:3000 ms | 記憶體限制:65535 KB 難度:4- 描述
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這有一個迷宮,有0~8行和0~8列:
1,1,1,1,1,1,1,1,1
1,0,0,1,0,0,1,0,1
1,0,0,1,1,0,0,0,1
1,0,1,0,1,1,0,1,1
1,0,0,0,0,1,0,0,1
1,1,0,1,0,1,0,0,1
1,1,0,1,0,1,0,0,1
1,1,0,1,0,0,0,0,1
1,1,1,1,1,1,1,1,10表示道路,1表示牆。
現在輸入一個道路的座標作為起點,再如輸入一個道路的座標作為終點,問最少走幾步才能從起點到達終點?
(注:一步是指從一座標點走到其上下左右相鄰座標點,如:從(3,1)到(4,1)。)
- 輸入
- 第一行輸入一個整數n(0<n<=100),表示有n組測試資料;
隨後n行,每行有四個整數a,b,c,d(0<=a,b,c,d<=8)分別表示起點的行、列,終點的行、列。 - 輸出
- 輸出最少走幾步。
- 樣例輸入
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2 3 1 5 7 3 1 6 7
- 樣例輸出
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12 11
水題一道,無腦廣搜就過了吧,以前寫的,沒什麼可說的,有點對不起這個難度,現在看用stl程式碼可以減少一半以上
#include<stdio.h> int map[9][9] = {1,1,1,1,1,1,1,1,1, 1,0,0,1,0,0,1,0,1, 1,0,0,1,1,0,0,0,1, 1,0,1,0,1,1,0,1,1, 1,0,0,0,0,1,0,0,1, 1,1,0,1,0,1,0,0,1, 1,1,0,1,0,1,0,0,1, 1,1,0,1,0,0,0,0,1, 1,1,1,1,1,1,1,1,1}; int queue[100][3] , haxi[9][9]; int head , tail; int i , start_x , start_y , end_x , end_y; void bfs() { while(head != tail) { if(queue[head][0] > 0 && map[queue[head][0] - 1][queue[head][1]] == 0 && haxi[queue[head][0] - 1][queue[head][1]] == 0) { queue[tail][0] = queue[head][0] - 1; queue[tail][1] = queue[head][1]; queue[tail][2] = queue[head][2] + 1; haxi[queue[tail][0]][queue[tail][1]] = 1; if(queue[tail][0] == end_x && queue[tail][1] == end_y) { printf("%d\n" , queue[tail][2]); return ; } tail++; } if(queue[head][0] < 8 && map[queue[head][0] + 1][queue[head][1]] == 0 && haxi[queue[head][0] + 1][queue[head][1]] == 0) { queue[tail][0] = queue[head][0] + 1; queue[tail][1] = queue[head][1]; queue[tail][2] = queue[head][2] + 1; haxi[queue[tail][0]][queue[tail][1]] = 1; if(queue[tail][0] == end_x && queue[tail][1] == end_y) { printf("%d\n" , queue[tail][2]); return ; } tail++; } if(queue[head][1] > 0 && map[queue[head][0]][queue[head][1] - 1] == 0 && haxi[queue[head][0]][queue[head][1] - 1] == 0) { queue[tail][0] = queue[head][0]; queue[tail][1] = queue[head][1] - 1; queue[tail][2] = queue[head][2] + 1; haxi[queue[tail][0]][queue[tail][1]] = 1; if(queue[tail][0] == end_x && queue[tail][1] == end_y) { printf("%d\n" , queue[tail][2]); return ; } tail++; } if(queue[head][1] < 8 && map[queue[head][0]][queue[head][1] + 1] == 0 && haxi[queue[head][0]][queue[head][1] + 1] == 0) { queue[tail][0] = queue[head][0]; queue[tail][1] = queue[head][1] + 1; queue[tail][2] = queue[head][2] + 1; haxi[queue[tail][0]][queue[tail][1]] = 1; if(queue[tail][0] == end_x && queue[tail][1] == end_y) { printf("%d\n" , queue[tail][2]); return ; } tail++; } head++; } } int find_way(int x , int y) { head = 0 ; tail = 1; queue[head][0] = start_x; queue[head][1] = start_y; haxi[queue[head][0]][queue[head][1]] = 1; bfs(); if(head == tail) return 0; else return 1; } int main() { int x,y; scanf("%d" , &i); while(i--) { scanf("%d %d %d %d" , &start_x , &start_y , &end_x , &end_y); for(x = 0 ; x < 9 ; x++) for(y = 0 ; y < 9 ; y++) haxi[x][y] = 0; if(find_way(start_x , start_y) == 0) printf("0\n"); } return 0; }