POJ 1925 Spiderman

题意：如下图所示，蜘蛛侠在某个位置，他女朋友在另一个建筑(WestTower)上，蜘蛛侠每次可以吐丝到附近的建筑顶端，然后摇摆到关于该建筑物对称的位置。只要荡到横坐标 >= X(WestTower)的位置，就能爬上WestTower。问在不碰到地的情况下，最少要吐多少次丝。

/*
* Author:  Plumrain
* Created Time:  2013-11-18 19:22
* File Name: DP-POJ-1925.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

#define CLR(x) memset(x, 0, sizeof(x))
const int maxint = 2147483647 - 10;
typedef long long int64;

int n;
int x[5005], y[5005], d[1010005];
int64 a[5005];

void init()
{
scanf ("%d", &n);
for (int i = 0; i < n; ++ i){
scanf ("%d%d", &x[i], &y[i]);
a[i] = (int64)y[i]*y[i] - (int64)(y[i]-y[0]) * (y[i]-y[0]);
}
}

int DP()
{
for (int i = 0; i <= x[n-1]+10000; ++ i)
d[i] = maxint;
d[x[0]] = 0;

for (int i = 1; i < n; ++ i){
for (int j = x[i]-1; j >= x[0]; -- j){
if ((int64)(x[i]-j)*(x[i]-j) > a[i]) break;

int tmp = 2*x[i] - j;
if (tmp >= x[n-1])
tmp = x[n-1];
d[tmp] = min(d[tmp], d[j] + 1);
}
}

return d[x[n-1]] == maxint ? -1 : d[x[n-1]];
}

int main()
{
int T;
scanf ("%d", &T);
while (T--){
init();
printf ("%d\n", DP());
}
return 0;
}

View Code