Hoj 2245 planktonic Triangular cell (mathematics AH)

Title Link: http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=2245

Time limit:500 MS Memory limit:50000 K Total submit:129(users) Total accepted:66(users) Rating: Special Judge: No

Description

, Zoidtrip is a casual to small game ... The player needs to manipulate a triangular cell that keeps moving forward at a speed of 45 degrees downward, constantly avoiding horizontal obstructions, and can change direction each time the screen is clicked. [Oblique left (right) 45° can be transformed to oblique right (left) 45°] Now, there are N-layer obstacles, layer I obstacles can be worn from the range of the horizontal axis l[i]~r[i] (including l[i] and R[i]), and the distance between the layer I obstacle and the i-1 barrier is d[i]. May I ask, assuming that you can move up to the first layer of the field if you can change direction infinitely? We stipulate that the player is born on the No. 0 floor, where the horizontal axis is 0. You can change direction at any real number moment.

Input

Multiple sets of test data. The first behavior of each set of test data is two positive integers n and v. Next n lines, 3 integers per line l[i], R[i], d[i]. (N <= 2000000,0 <= all data < 2^31)

Output

For each set of data, the output line contains an integer that represents the maximum number of layers to advance to.

Sample Input

3 7 1 3 1 4 10 5 8 10 1 4 1 1 1 1 2 5 10 1 1 1 3 5 2

Sample Output

2 4

Hint

"The distance between the layer I obstacle and the barrier of the i-1 layer is d[i]" So d[1] is the distance between the first and 0th floors. Sample 1 is explained as follows: We can move from the birth position to the right down to the first layer with coordinates 1. Next, you can continue to move right down to the second-level coordinate of 6. But in any case it cannot be moved to the third level between the 8~10. Example 2 illustrates the following: (0,0), (2,2), (3,1), (4,3) So it reaches the fourth floor.

Source

The fifth annual ACM Program Design Contest of Harbin Polytechnic University

Ps:

Find the triangular cell at the left and right distance from each layer that can be reached and meet the obstacles.

The code is as follows:

#include <cstdio> #include <iostream> #include <algorithm>using namespace std; #define LL Long long# Define MAXN 2000047LL L[MAXN], R[MAXN], D[maxn];int main () {    LL n, v;    while (scanf ("%lld%lld", &n,&v)!=eof)    {        LL L = 0,r = 0;        int ans = 0;        for (int i=0; i<n; i++)        {            scanf ("%lld%lld%lld", &l[i],&r[i],&d[i]);        }        for (int i = 0; i < n; i++)        {            if (L[i] > R[i])            {                LL t = r[i];                R[i] = L[i];                L[i] = t;            }            L-=d[i];            R+=d[i];            L = max (l[i],l);            R = min (r[i],r);            if (L > R)            {break                ;            }            ans++;        }        if (v = = 0)            ans = 0;        printf ("%d\n", ans);    }    return 0;}

Hoj 2245 planktonic Triangular cell (mathematics AH)