第一题：染色问题（NOIP2009模拟测试题）

染色问题 

问题描述：

有一段从0到1000000000的数轴，它开始的颜色是白色。现在有人不断把其中的一段染成黑色或白色，总共染了N段（1<=N<=5000）。你的任务是编写一个程序，找出最后最长的白色段。

输入：第一行只有一个数N，接下来的N行是每次染一段的信息，格式为：ai bi ci。

ai、bi是整数，ci是符号’b’或’w’，三者用空格隔开，表示这次从ai染到bi，用的颜色为ci（’w’表示白色，’b’表示黑色），你可以认为0<ai<=bi<1000000000。

输出：仅两个数x，y（x<=y），用空格隔开，表示最长的白色段。如果有多个解，则输出X最小的解。

〖输入输出样例〗：

dye.in

dye.out

4

1 999999997 b

40 300 w

300 634 w

43 47 b

 

48 634

 

/****************************************************************************************************** ** Copyright (C) 2011.07.01-2013.07.01 ** Author: famousDT <[e-mail]> ** Edit date: 2011-10-17******************************************************************************************************/#include <stdio.h>#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10#include <vector>#include <queue>#include <map>#include <time.h>#include <set>#include <list>#include <stack>#include <string>#include <iostream>#include <assert.h>#include <string.h>//memcpy(to,from,count#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll#include <algorithm>using namespace std;//typedef long long ll;#define MY_PI acos(-1)#define MY_MAX(a, b) ((a) > (b) ? (a) : (b))#define MY_MIN(a, b) ((a) < (b) ? (a) : (b))#define MY_MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))#define MY_ABS(a) (((a) >= 0) ? (a) : (-(a)))#define MY_INT_MAX 0x7fffffff/*==========================================================*\| noip\*==========================================================*/struct segment{    int low;    int high;} seg[100005], tmp[100005];bool cmp(segment a, segment b){    return a.low < b.low;}int main(){    FILE *in, *out;    in = freopen("dye.in", "r", stdin);    out = freopen("dye.out", "w", stdout);int n;int i, j, k, a, b;int index = 1;char c;scanf("%d", &n);seg[0].low = 0;seg[0].high = 1000000000;for (k = 0; k < n; ++k) {scanf("%d%d %c", &a, &b, &c);if (c == 'w') {//合并白色区间，可以用二分int pos = 0;seg[index].low = a;seg[index].high = b;++index;sort(seg, seg + index, cmp);int t_low = seg[0].low;int t_high = seg[0].high;for (i = 0; i < index; ++i) {if (seg[i].low <= t_high) {if (seg[i].high > t_high) {t_high = seg[i].high;}} else {seg[pos].low = t_low;seg[pos++].high = t_high;t_low = seg[i].low;t_high = seg[i].high;}}seg[pos].low = t_low;seg[pos++].high = t_high;index = pos;        } else {//白色区间减去黑色区间，分情况讨论int pos = 0;for (i = 0; i < index; ++i) {if (seg[i].high < a || seg[i].low > b) {tmp[pos].low = seg[i].low;tmp[pos].high = seg[i].high;++pos;} else if (seg[i].low >= a && seg[i].high <= b) {;} else if (seg[i].low >= a && seg[i].high > b) {tmp[pos].low = b + 1;tmp[pos].high = seg[i].high;++pos;} else if (seg[i].low < a && seg[i].high <= b) {tmp[pos].low = seg[i].low;tmp[pos].high = a - 1;++pos;} else if (seg[i].low < a && seg[i].high > b) {tmp[pos].low = seg[i].low;tmp[pos].high = a - 1;++pos;tmp[pos].low = b + 1;tmp[pos].high = seg[i].high;++pos;}}for (i = 0; i < pos; ++i) {seg[i].low = tmp[i].low;seg[i].high = tmp[i].high;}index = pos;}    }int max = -1;int ans;for (i = 0; i < index; ++i) {int t = seg[i].high - seg[i].low;if (t > max) {max = t;ans = i;}}printf("%d %d\n", seg[ans].low, seg[ans].high);system("pause");return 0;}/*41 999999997 b40 300 w300 634 w43 47 b*/