codeforces 589d（相遇问题）

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题目链接：http://codeforces.com/problemset/problem/589/D

题意：给出n个人行走的开始时间和结束时间，求每个人分别能跟多少人相遇打招呼。

分析：两个人行走无非四种情况，同向两种：-> -> 和 <- <-反向两种-> <- 和<- ->分情况推他们相遇的条件即可，我是列方程求解的。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<map>#include<vector>#include<cmath>#include<set>#include<queue>using namespace std;typedef long long ll;const int maxn = 1000 + 10;const int maxt = 1e6 + 10;const double eps = 0.00000000000001;const int inf = 0x3f3f3f3f;struct person{    int num,t,b,e,time;}p[maxn];int hellow[maxn];bool check1(int a,int b, int c, int d)//看区间[a,b]与区间[c,d]是否有交集{    if(c <= a && b <= d) return true;    if(a <= c && c <= b) return true;    if(a <= d && d <= b) return true;    return false;}bool check2(double t1, double t11, double t2, double t22, double t)//判断t是否同时在区间[t1,t2]和区间[t11,t22]内{    return t >= t1 && t <= t11 && t >= t2 && t <= t22;}int main(){    int k;    while(~scanf("%d",&k))    {        memset(hellow, 0, sizeof(hellow));        memset(p, 0, sizeof(p));        for(int i = 0; i < k; i++)        {            scanf("%d%d%d",&p[i].t, &p[i].b, &p[i].e);        }        for(int i = 0; i < k; i++)//两两看是否相遇        {            for(int j = i + 1; j < k; j++)            {                int t1 = p[i].t, t11 = p[i].t + abs(p[i].e - p[i].b);//t1:第i个人开始的时间  t11:第i个人结束的时间                int t2 = p[j].t, t22 = p[j].t + abs(p[j].e - p[j].b);//同上                int fi = p[i].b <= p[i].e ? 1 : -1;                int fj = p[j].b <= p[j].e ? 1 : -1;                int b1 = p[i].b, e1 = p[i].e, b2 = p[j].b, e2 = p[j].e;                if(fi == 1 && fj == 1)//->  ->                {                    if(t1 - b1 == t2 - b2 && check1(t1,t11,t2,t22))                    {                        hellow[i]++,hellow[j]++;                    }                }                if(fi == 1 && fj == -1)//-> <-                {                    double t = (double)(t1 - b1 + t2 + b2);                    t /= 2;                    if(check2(t1,t11,t2,t22,t))                    {                        hellow[i]++,hellow[j]++;                    }                }                if(fi == -1 && fj == 1)//<- ->                {                    double t = (double)(t1 + b1 + t2 - b2);                    t /= 2;                    if(check2(t1,t11,t2,t22,t))                    {                        hellow[i]++,hellow[j]++;                    }                }                if(fi == -1 && fj == -1)//<-  <-                {                    if(t1 + b1 == t2 + b2 &&  check1(t1,t11,t2,t22))                    {                        hellow[i]++,hellow[j]++;                    }                }            }        }        for(int i = 0; i < k; i++)        {            if(i) printf(" ");            printf("%d",hellow[i]);        }        printf("\n");    }    return 0;}

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