Right turn

最新推荐文章于 2021-01-29 04:28:10 发布

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最新推荐文章于 2021-01-29 04:28:10 发布

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原文链接：http://www.cnblogs.com/ACMessi/p/4852278.html

Right turn

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

frog is trapped in a maze. The maze is infinitely large and divided into grids. It also consists of

frog is initially in grid

The maze is so large that frog has no chance to escape. Help her find out the number of turns she will make.

Input

The input consists of multiple tests. For each test:

The first line contains

Output

For each test, write

Sample Input

Sample Output

2
0
-1

4种情况的dfs:

　　　　只要把所有情况用代码表达清楚就OK了。这里有几点较为关键：dfs()该传递什么值，如何判断重复，如何找到运动时遇到的第一个障碍物。

　　　　首先，dfs()传递的a[].x,a[].y是障碍物的位置，而物体实际的位置，应该是(a[].x,a[].y-1),(a[].x-1,a[].y),(a[].x,a[].y+1),(a[].x+1,a[].y)这四个不同的状态;所以在判断turn%4后，要先处理一下x和y;所以最初放入dfs()的并不应该是(0,0)，而是(0,1)。

　　　　撞到同一个障碍物只有4种方向，并且这4种直接可以用turn%4来表示，判断dis[][]如若有相同，则必有重复。

　　　　turn right 的要求不过是遇到一个同x(or y)的比当前位置y(or x) 大(or 小) 的最小(or 最大) 值，找到则turn++,dfs();否则结束。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
const int INF = 0x3f3f3f3f;
using namespace std;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,-1,0,1};
int dis[1005][4];
int sign;
int turn;
int n;
struct node{
    int x,y;
}a[1005];
void dfs(int cnt){
    if(sign)return;
    int tu = turn%4;
    for(int i = 0; i < 4; i++){
        if(dis[cnt][i] == tu){
            sign = 2;return;
        }
        if(dis[cnt][i] == -1){
            dis[cnt][i] = tu;
            break;
        }
    }
    int j = -1;
    if(dy[tu] == 0){
        if(dx[tu] == 1){
            int x = a[cnt].x;
            int y = a[cnt].y-1;
            int xx = INF;
            for(int i = 1; i <= n; i++){
                if(a[i].x > x && a[i].x < xx && a[i].y == y){
                    xx = a[i].x;
                    j = i;
                }
            }
            if(j == -1){
                sign = 1;
                return;
            }
            else {
                turn++;
                dfs(j);
            }
        }
        else {
            int x = a[cnt].x;
            int y = a[cnt].y+1;
            int xx = -INF;
            for(int i = 1; i <= n; i++){
                if(a[i].x < x && a[i].x > xx && a[i].y == y){
                    xx = a[i].x;
                    j = i;
                }
            }
            if(j == -1){
                sign = 1;
                return;
            }
            else {
                turn++;
                dfs(j);
            }
        }
    }
    else {
        if(dy[tu] == -1){
            int x = a[cnt].x-1;
            int y = a[cnt].y;
            int yy = -INF;
            for(int i = 1; i <= n; i++){
                if(a[i].y < y && a[i].y > yy && a[i].x == x){
                    j = i;
                    yy = a[i].y;
                }
            }
            if(j == -1){
                sign = 1;
                return;
            }
            else {
                turn++;
                dfs(j);
            }
        }
        else {
            int x = a[cnt].x+1;
            int y = a[cnt].y;
            int yy = INF;
            for(int i = 1; i <= n; i++){
                if(a[i].y > y && a[i].y < yy && a[i].x == x){
                    yy = a[i].y;
                    j = i;
                }
            }
            if(j == -1){
                sign = 1;
                return;
            }
            else {
                turn++;
                dfs(j);
            }
        }
    }
    return;
}
int main(){
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n; i++){
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        memset(dis,-1,sizeof(dis));
        sign = 0;
        turn = 0;
        a[0].x = 0,a[0].y = 1;
        dfs(0);
        if(sign == 2)puts("-1");
        else printf("%d\n",turn);
    }
    return 0;
}