HDU1199-Color the Ball

Color the Ball

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                Total Submission(s): 6032    Accepted Submission(s): 1507

Problem Description

There 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.

 

Input

First 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.

 

Output

Two 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 Input

31 4 w8 11 w3 5 b

 

Sample Output

8 11

 

Author

ZHOU, Kai

 

Source

ZOJ Monthly, February 2005 

 

Recommend

Ignatius.L

 

题意：有若干个球以及n次操作，每次操作把区间 [ a , b ] 的球涂成一种颜色，w 表示涂成白色，b 表示涂成黑色。初始的时候所有的球都是黑色的。求出n次操作以后最长连续白色的区间，如果有多个，输出最左边的那个

解题思路：线段树+离散化（注意离散化时需要将一个点的左右两个点也加进去）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int lr[4 * 12007], rl[4 * 12007], ma[4 * 12007], lazy[4 * 12007];int xx[12007], n, cnt;int L[2005], R[2005];char ch[2005][3];map<int, int>mp;void Merge(int k, int l, int r){lr[k] = lr[k << 1], rl[k] = rl[k << 1 | 1];ma[k] = max(ma[k << 1], ma[k << 1 | 1]);ma[k] = max(ma[k], rl[k << 1] + lr[k << 1 | 1]);int mid = (l + r) / 2;int L = xx[mid] - xx[l - 1];int R = xx[r] - xx[mid];if (lr[k << 1] == L) lr[k] += lr[k << 1 | 1];if (rl[k << 1 | 1] == R) rl[k] += rl[k << 1];}void Push(int k, int l, int r){lazy[k << 1] = lazy[k << 1 | 1] = lazy[k];int mid = (l + r) / 2;int L = xx[mid] - xx[l - 1];int R = xx[r] - xx[mid];ma[k << 1] = lr[k << 1] = rl[k << 1] = L*lazy[k];ma[k << 1 | 1] = lr[k << 1 | 1] = rl[k << 1 | 1] = R*lazy[k];lazy[k] = -1;}void update(int k, int l, int r, int ll, int rr, int flag){if (ll <= l && rr >= r){lazy[k] = flag;ma[k] = lr[k] = rl[k] = flag * (xx[r] - xx[l - 1]);return;}if (lazy[k] != -1) Push(k, l, r);int mid = (l + r) / 2;if (ll <= mid) update(k << 1, l, mid, ll, rr, flag);if (rr > mid) update(k << 1 | 1, mid + 1, r, ll, rr, flag);Merge(k, l, r);}int query(int k, int l, int r){int mid = (l + r) / 2;if (lazy[k] != -1) Push(k, l, r);if (ma[k << 1] == ma[1]) return query(k << 1, l, mid);else if (rl[k << 1] + lr[k << 1 | 1] == ma[1]) return xx[mid] - rl[k << 1] + 1;else return query(k << 1 | 1, mid + 1, r);}int main(){while (~scanf("%d", &n)){mp.clear();cnt = 1;for (int i = 1; i <= n; i++){scanf("%d%d%s", &L[i], &R[i], ch[i]);xx[cnt++] = L[i] - 1, xx[cnt++] = L[i], xx[cnt++] = L[i] + 1;xx[cnt++] = R[i] - 1, xx[cnt++] = R[i], xx[cnt++] = R[i] + 1;}sort(xx + 1, xx + cnt);cnt = unique(xx + 1, xx + cnt) - xx;memset(ma, 0, sizeof ma);memset(lr, 0, sizeof lr);memset(rl, 0, sizeof rl);memset(lazy, -1, sizeof lazy);for (int i = 1; i < cnt; i++) mp[xx[i]] = i;for (int i = 1; i <= n; i++){int a = mp[L[i]], b = mp[R[i]];if (ch[i][0] == 'w') update(1, 1, cnt - 1, a, b, 1);else update(1, 1, cnt - 1, a, b, 0);}if (ma[1] <= 0) puts("Oh, my god");else{int l = query(1, 1, cnt - 1);printf("%d %d\n", l, l+ma[1]-1);}}return 0;}