最少步数
时间限制:3000 ms | 内存限制:65535 KB
难度:4
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描述
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这有一个迷宫,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表示墙。
现在输入一个道路的坐标作为起点,再如输入一个道路的坐标作为终点,问最少走几步才能从起点到达终点?
(注:一步是指从一坐标点走到其上下左右相邻坐标点,如:从(3,1)到(4,1)。)
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输入
- 第一行输入一个整数n(0<n<=100),表示有n组测试数据;
随后n行,每行有四个整数a,b,c,d(0<=a,b,c,d<=8)分别表示起点的行、列,终点的行、列。
输出 - 输出最少走几步。 样例输入
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2 3 1 5 7 3 1 6 7
样例输出 -
12 11
- 第一行输入一个整数n(0<n<=100),表示有n组测试数据;
典型的bfs搜索问题
#include<cstdio>
#include<queue>
#include<iostream>
using namespace std;
typedef pair<int, int> P;
const int INF = 100000000;
int d[9][9];
int a[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 dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int sx, sy, gx, gy, n;
int bfs() {
queue<P> que;
for (int i = 0; i < 9; i++)
for (int j = 0; j < 9; j++)
d[i][j] = INF;
que.push(P(sx, sy));
d[sx][sy] = 0;
while(que.size()) {
//puts("aa");
P p = que.front();
que.pop();
if (p.first == gx && p.second == gy) break;
// 四个方向的循环
for (int i = 0; i < 4; i++) {
int nx = p.first + dx[i], ny = p.second + dy[i];
if (nx >= 1 && nx < 9 && ny >= 1 && ny < 9 && a[nx][ny] != 1 && d[nx][ny] == INF) {
que.push(P(nx, ny));
d[nx][ny] = d[p.first][p.second] + 1;
}
}
}
return d[gx][gy];
}
int main() {
//freopen("in.txt", "r", stdin);
scanf("%d", &n);
while(n--) {
scanf("%d%d%d%d", &sx, &sy, &gx, &gy);
int res = bfs();
printf("%d\n", res);
}
return 0;
}