Gym 100531J Joy of Flight (几何)

题意：你从开始坐标到末尾坐标，要经过 k 秒，然后给你每秒的风向，和飞机的最大速度，问能不能从开始到末尾。

析：首先这个风向是不确定的，所以我们先排除风向的影响，然后算出，静风是的最小速度，如果这都大于最大速度，肯定是不可能，如果可能，

再计算出每秒走的单位长度，然后再模拟整个过程。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const LL mod = 1e3 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int t[maxn];
double x[maxn], y[maxn];

int main(){
    freopen("joy.in", "r", stdin);
    freopen("joy.out", "w", stdout);
    int sx, sy, fx, fy;
    while(scanf("%d %d %d %d", &sx, &sy, &fx, &fy) == 4){
        int k;
        scanf("%d %d %d", &n, &k, &m);
        for(int i = 1; i <= n; ++i)  scanf("%d %lf %lf", &t[i], &x[i], &y[i]);
        t[n+1] = k;
        double dx = fx - sx;
        double dy = fy - sy;
        for(int i = 1; i <= n; ++i){
            dx -= x[i] * (t[i+1] - t[i]);
            dy -= y[i] * (t[i+1] - t[i]);
        }
        if(hypot(dx, dy) > (double)m * k){
            puts("No");  continue;
        }
        puts("Yes");
        double vx = dx / k;
        double vy = dy / k;
        int pos = 0;
        double ansx = sx;
        double ansy = sy;
        for(int i = 1; i <= k; ++i){
            if(t[pos+1] < i) ++pos;
            ansx += x[pos] + vx;
            ansy += y[pos] + vy;
            printf("%f %f
", ansx, ansy);
        }
    }
    return 0;
}