PKU 1925 - Spiderman (DP)
题意
蜘蛛侠要飞到最后一栋房子上。问最少需要几步。
思路
根据坐标来dp。
dp[i]
表示到i坐标需要的最少步数。
先处理出能连上某个楼的X的最长距离。
然后荡到
building[i].X+j
处，更新此处的距离。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int MAXN = 1e6 + 10;
const int MOD = 9901;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
int dp[MAXN];
pair<int, LL> arr[5005];
double scale[5005];
int main()
{
//ROP;
int T;
scanf("%d", &T);
while (T--)
{
MS(dp, -1);
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d%d", &arr[i].X, &arr[i].Y);
scale[i] = sqrt(arr[i].Y*arr[i].Y - (arr[i].Y-arr[1].Y)*(arr[i].Y-arr[1].Y));
}
dp[arr[1].X] = 0;
for (int i = 2; i <= n; i++)
for (int j = 1; j <= scale[i]; j++)
{
int tmp = arr[i].X-j;
if (tmp < arr[1].X) break;
if (dp[tmp] == -1) continue;
int aim = min(arr[i].X+j, arr[n].X);
dp[aim] = dp[aim] == -1 ? dp[tmp]+1 : min(dp[tmp]+1, dp[aim]);
}
printf("%d\n", dp[arr[n].X]);
}
return 0;
}