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Color the BallTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4028 Accepted Submission(s): 999Problem DescriptionThere are infinite balls in a line (numbered 1 2 3 ....), and initially all of them are paint black. Now Jim use a brush paint the balls, every time
give two integers a b and follow by a char
‘w‘ or ‘b‘, ‘w‘
denotes the ball from a to b are painted white, ‘b‘
denotes that be painted black. You are ask to find the longest white ball sequence. InputFirst line is an integer N (<=2000), the times Jim paint, next N line contain a b c, c can be ‘w‘
and ‘b‘.There are multiple cases, process to the end of file. OutputTwo integers the left end of the longest white ball sequence and the right end of longest white ball sequence (If more than one output the small number one). All the input are less than 2^31-1. If no such sequence exists, output "Oh, my god". Sample Input31 4 w8 11 w3 5 b Sample Output8 11 |
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<algorithm> #include<cstring> using namespace std; struct node{ int beg, en; }s[8000]; int len;//维持线段长度 bool cmp(node a, node b){ if (a.beg != b.beg) return a.beg < b.beg;//从小到大排有利于寻找可连接的段 else return a.en>b.en; } void Process(int h, int e, char c){ //只对将白涂黑进行处理，总共有4种情况 //离散区间为[ ] if (c == ‘w‘){ s[len].beg = h; s[len++].en = e; } else{ for (int i = 0; i < len; i++){ if (s[i].beg >= h&&s[i].en <= e){ s[i].beg = s[i].en = -1; } else if (s[i].beg >= h&&s[i].en >= e&&s[i].beg <= e){ s[i].beg = e + 1; } else if (s[i].beg <= h&&s[i].en >= e){ s[len].en = s[i].en;//先这一步，否则会被覆盖，^v^ s[i].en = h - 1; s[len++].beg = e + 1; } else if (s[i].beg <= h&&s[i].en >= h&&s[i].en<=e){ s[i].en = h - 1; } } } } int main() { int cnt, n, h, e, x, y; char c; while (scanf("%d", &n) != -1){ memset(s, 0, sizeof(s)); len = 0; for (int i = 0; i < n; i++){ scanf("%d %d %c", &h, &e, &c); Process(h, e, c); } cnt = h = e = -1; sort(s, s + len, cmp); for (int i = 0; i < len; i++){ if (s[i].beg != -1){ if (s[i].beg <= e + 1){ e = max(e, s[i].en); } else{ h = s[i].beg; e = s[i].en; } if (cnt < e - h + 1){ cnt = e - h + 1; x = h; y = e; } } } if (cnt == -1){ printf("Oh, my god\n"); } else{ printf("%d %d\n", x, y); } } return 0; }
Color the Ball2,布布扣,bubuko.com
原文：http://www.cnblogs.com/littlehoom/p/3569904.html