CF 237

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http://codeforces.com/contest/404/problem/B

给一个a*a的正方形，起点在（0,0）,每隔d米有一处提供水的地方，给出询问n，问前n处提供水的地点的坐标。

简单的模拟题，由于做的时间太长，导致C题没来得及做。

By 12120501045, contest: Codeforces Round #237 (Div. 2), problem: (B) Marathon, Accepted, ##include <stdio.h>#include <string.h>#include <algorithm>#include <cmath>using namespace std;int main(){    double a,d;    int n;    while(~scanf("%lf %lf",&a,&d))    {        scanf("%d",&n);        double x = 0, y = 0;        double L = 4*a;        double tmp;        int t;        double res;        double sh;        for(int k = 1; k <= n; k++)        {            if( y == 0)            {                tmp = a-x;                if(tmp >= d)                {                    printf("%.10lf 0.0000000000\n",x+d);                    x = x+d;                    y = 0;                }                else                {                    res = d-tmp;                    t = res/L;                    sh = res-t*L;                    if(sh <= a)                    {                        x = a;y = sh;                        printf("%.10lf %.10lf\n",x,y);                    }                    else if(sh > a && sh <= 2*a)                    {                        x = 2*a-sh;y = a;                        printf("%.10lf %.10lf\n",x,y);                    }                    else if(sh > 2*a && sh <= 3*a)                    {                        x = 0;y = 3*a-sh;                        printf("0.0000000000 %.10lf\n",y);                    }                    else if(sh > 3*a && sh <= 4*a)                    {                        x = sh-3*a;y = 0;                        printf("%.10lf 0.0000000000\n",x);                    }                }            }            else if(x == a)            {                tmp = a-y;                if(tmp >= d)                {                    printf("%.10lf %.10lf\n",a,y+d);                    x = a;y = y+d;                }                else                {                    double res = d-tmp;                    int t = res/L;                    double sh = res-t*L;                    if(sh <= a)                    {                        x = a-sh; y = a;                        printf("%.10lf %.10lf\n",x,y);                    }                    else if(sh > a && sh <= 2*a)                    {                        x = 0; y = 2*a-sh;                        printf("0.0000000000 %.10lf\n",y);                    }                    else if(sh > 2*a && sh <= 3*a)                    {                        x = sh-2*a; y = 0;                        printf("%.10lf 0.0000000000\n",x);                    }                    else if(sh > 3*a && sh <= 4*a)                    {                        x = a;y = sh-3*a;                        printf("%.10lf %.10lf\n",x,y);                    }                }            }            else if(y == a)            {                tmp = x;                if(tmp >= d)                {                    printf("%.10lf %.10lf\n",x-d,a);                    x = x-d;y = a;                }                else                {                    double res = d-tmp;                    int t = res/L;                    double sh = res-t*L;                    if(sh <= a)                    {                        x = 0; y = a-sh;                        printf("0.0000000000 %.10lf\n",y);                    }                    else if(sh > a && sh <= 2*a)                    {                        x = sh-a; y = 0;                        printf("%.10lf 0.0000000000\n",x);                    }                    else if(sh > 2*a && sh <= 3*a)                    {                        x = a; y = sh-2*a;                        printf("%.10lf %.10lf\n",x,y);                    }                    else if(sh > 3*a && sh <= 4*a)                    {                        x = 4*a-sh; y = a;                        printf("%.10lf %.10lf\n",x,y);                    }                }            }            else if(x == 0)            {                tmp = y;                if(tmp >= d)                {                    printf("0.0000000000 %.10lf\n",tmp-d);                    x = 0; y = tmp-d;                }                else                {                    double res = d-tmp;                    int t = res/L;                    double sh = res-t*L;                    if(sh <= a)                    {                        x = sh; y = 0;                        printf("%.10lf 0.0000000000\n",x);                    }                    else if(sh > a && sh <= 2*a)                    {                        x = a; y = sh-a;                        printf("%.10lf %.10lf\n",x,y);                    }                    else if(sh > 2*a && sh <= 3*a)                    {                        x = 3*a-sh; y = a;                        printf("%.10lf %.10lf\n",x,y);                    }                    else if(sh > 3*a && sh <= 4*a)                    {                        x = 0; y = 4*a-sh;                        printf("0.0000000000 %.10lf\n",y);                    }                }            }        }    }    return 0;}

http://codeforces.com/problemset/problem/404/C

题意：给出n个点和每个点最多连的边数k。接下来输入每个点到某一个点的最短距离（两点相连距离为1），构建一个无向图，满足上述距离。

注意到距离为0的点只能有一个，那边是源点。这个图的边数最少是n-1条边，然后按距离源点从小到大加边就OK。

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>using namespace std;const int INF = 0x3f3f3f3f;vector <int> dis[100010]; //dis[i]存距离源点为i的点vector <pair <int,int> > ans;int n,k;int maxdis;int main(){while(~scanf("%d %d",&n,&k)){maxdis = -1;int x;for(int i = 0; i <= n; i++){dis[i].clear();ans.clear();}for(int i = 1; i <= n; i++){scanf("%d",&x);maxdis = max(maxdis,x);dis[x].push_back(i);}if(dis[0].size()!= 1){printf("-1\n");return 0;}for(int i = 1; i <= maxdis; i++){int index = 0;//初始化，index为距离是i-1的第一个点int cnt = (i!=1); //cnt记录index发出去的边数，i!=1时为1,。for(int j = 0; j < (int)dis[i].size(); j++){if(cnt == k){index++;cnt = (i!=1);}if(index == (int)dis[i-1].size()){printf("-1\n");return 0;}ans.push_back( make_pair(dis[i-1][index], dis[i][j]));cnt++;}}printf("%d\n",n-1);for(int i = 0; i < (int)ans.size(); i++)printf("%d %d\n",ans[i].first, ans[i].second);}return 0;}

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