POJ - 1054 The Troublesome Frog

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题意：给你个矩阵，里面有n个标记的点，许多只青蛙在上面跳，每次跳的距离都是一样的且轨迹是直线，目标是从一边跳到另一边，求最多步数的青蛙

思路：排序后，枚举判断

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 5050;

struct point{
	int x,y;
	void init(int X, int Y) {
		x = X;
		y = Y;
	}
}a[MAXN];
int map[MAXN][MAXN];
int r,c,n;

int cmp(point A, point B) {
	if (A.x == B.x)
		return A.y < B.y;
	return A.x < B.x;
}

int in(int x, int y) {
	if (x < 1 || y < 1 || x > r || y > c)
		return 0;
	return 1;
}

int main() {
	int x,y;
	scanf("%d%d", &r ,&c);
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d%d", &a[i].x, &a[i].y);
		map[a[i].x][a[i].y] = 1;
	}
	sort(a, a+n, cmp);
	int ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = i+1; j < n; j++) {
			int nx = a[j].x - a[i].x;
			int ny = a[j].y - a[i].y;
			if (in(a[i].x-nx, a[i].y-ny) || !in(a[j].x+ans*nx, a[j].y+ans*ny) || !map[a[j].x+ans*nx][a[j].y+ans*ny])
				continue;
			int cnt = 0;
			int x = a[j].x + nx;
			int y = a[j].y + ny;
			int flag = 1;
			while (in(x, y)) {
				if (!map[x][y]){
					flag = 0;
					break;
				}
				else cnt++;
				x += nx;
				y += ny;
			}
			if (flag) 
				ans = cnt;
		}
	}
	printf("%d\n", ans>0?ans+2:0);
	return 0;
}

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POJ - 1054 The Troublesome Frog

标签：style   class   blog   c   code   a   

原文地址：http://blog.csdn.net/u011345136/article/details/26478517