ZOJ 2763 Prison Break

题意：一群人要越狱，要用绳子荡过一条沟，给你初速度和体重，和对岸墙的高度，先做单摆运动后是斜抛，当高度和墙同高时松手斜抛。问你最后能成功越狱的有谁名字按字典序输出。

#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
#include <cstdio>
#include <math.h>
#include <vector>
using namespace std;
#define inf 1000000000
#define eps 1e-8
#define G 9.8

int dd(double x,double y){ return fabs(x-y)<eps;} // x==y
int dy(double x,double y){ return x>y+eps;}   // x>y
int xy(double x,double y){ return x<y-eps;}   //x<y
int dyd(double x,double y){ return x>y-eps;}  //x>=y
int xyd(double x,double y){ return x<y+eps;}  //x<=y

int main() {
    int m;
    double L,W,H,v,w,vx,vy,wei;
    string name;
    while(scanf("%d",&m) == 1) {
        vector<string> ans(0);
        scanf("%lf%lf%lf",&L,&W,&H);
        while(m--) {
            cin >> name;
            scanf("%lf%lf",&wei,&v);
            if(dy( H, L)) 
                continue;
            w = sqrt(H*(2.0*L-H));
            if(dy( w, W)) //撞墙了！
                continue;
            if(dy( 2.0*G*H, v*v))//到不了这高度
                continue;
            v = sqrt(fabs(v*v - 2.0*G*H));
            double vx = v * (L-H) / L;
            double vy = v * w / L;
            w += 2.0*vx * vy / G;
            if(xy( w, W)) 
                continue;
            ans.push_back(name);
        }
        sort(ans.begin(), ans.end());
        cout << ans.size() << endl;
        for(int i = 0; i < ans.size(); ++i)
            cout << ans[i] << endl;
    }
    return 0;
}

注：vector的快速排序，和精度处理