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; }
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; }