Milking Cows

题目：http://ace.delos.com/usacoprob2?S=milk2&a=b69TBYTWb4u

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题目大意：有N个时间区间 合并后 找出最大的连续区间 和 空闲区间

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简要分析：区间合并  1.先按照每个区间的起始值升序排列所有区间 2.从i=1,j=2合并 3.如果可以合并则把i的区间弄成（0，0） 4.i=j，j++ 5.当j等于N时 则所有区间都合并 6.再次排序区间 则（0，0）的区间都在前面 真正合并后的区间在后面

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/*ID: KJlmfePROG: milk2LANG: C++*/#include <stdio.h>FILE *in,*out;int n, low[5002], high[5002], MaxMilk = 0, MaxRest = 0;void Input(){ fscanf(in,"%d",&n); for(int i=1; i<=n; i++) fscanf(in,"%d %d",&low[i],&high[i]); }void Sort(){ for(int i=1; i<=n; i++) for(int j=1; j<=n-i; j++) if(low[j] > low[j+1]) { low[0] = low[j]; low[j] = low[j+1]; low[j+1] = low[0]; high[0] = high[j]; high[j] = high[j+1]; high[j+1] = high[0]; }}void Merge(){ int i = 1; int j = 2; while(i<=n && j<=n) if(low[j]>high[i]) { i=j; j++; } else { low[j] = low[j] < low[i] ? low[j] : low[i]; high[j] = high[j] > high[i] ? high[j] : high[i]; low[i] = 0; high[i] = 0; i = j; j++; } }void FindAnswer(){ int i; for(i=1;i<=n;i++) if(low[i]!=0 || high[i]!=0) break; for( ; i<=n; i++) { MaxMilk = MaxMilk > (high[i] - low[i]) ? MaxMilk : (high[i] - low[i]); if(i!=n) MaxRest = MaxRest > (low[i+1] - high[i]) ? MaxRest : (low[i+1] - high[i]); }}void Output(){ fprintf(out,"%d %d/n",MaxMilk,MaxRest); }int main(){ in = fopen("milk2.in","r"); out = fopen("milk2.out","w"); Input(); Sort(); Merge(); Sort(); FindAnswer(); Output(); fclose(in); fclose(out); return 0; }