Milking Cows(USACO)

/* ID:tianlin2 PROG:milk2 LANG:C++ */ #include <iostream> #include <fstream> using namespace std; int main() { ofstream fout("milk2.out"); ifstream fin("milk2.in"); int be1[5000],end1[5000]; //a为奶牛数,b为不挤奶的时间,c为最长不挤奶的时间,x为至少一个人挤奶的时间,y为最长挤奶时间 int a,b=0,c=0,x=0,y=0; fin>>a; for(int i=0;i!=a;++i) fin>>be1[i]>>end1[i]; //用一个数组做就可以了, //int *be1=new int[a]; //int *end1=new int[a]; /*for(int i=0;i!=a;++i) { be1[i]=be[i]; end1[i]=end[i]; }*/ //用数轴来考虑,分四种出现重复时间的情况,合并 for(int i=0;i!=a;++i) { for(int j=0;j!=a;++j) { if(end1[j]>=be1[i]&&end1[j]<=end1[i]&&be1[j]<=be1[i]) { be1[i]=be1[j]; end1[j]=end1[i]; } else if(be1[j]>=be1[i]&&end1[j]<=be1[i]) { be1[j]=be1[i]; end1[j]=end1[i]; } else if(be1[j]<=be1[i]&&end1[j]>=end1[i]) { be1[i]=be1[j]; end1[i]=end1[j]; } else if(be1[j]<=end1[i]&&be1[j]>=be1[i]&&end1[j]>=end1[i]) { be1[j]=be1[i]; end1[i]=end1[j]; } else continue; } } //以上处理完后仍然有,重复的,重复情况为第二种或者第三种 //算最大不挤奶时间错误的算法 /* for(int i=0;i!=a;++i) { b=0; int d=0; //MAX for(int j=0;j!=a;++j) { if(be1[i]>end1[j]) b=be1[i]-end1[j]; if(end1[i]<be1[j]) continue; if(be1[i]==be1[j]&&end1[i]==end1[j]) continue; if(d==0&&b!=0) d=b; //相当于D的初始化 if(b<d) d=b; if(b==8&&u==0&&d==8) { cout<<be1[i]<<' '<<end1[i]<<endl; cout<<be1[j]<<' '<<end1[j]<<endl; ++u; } } if(d>c) c=d; }*/ for(int i=0;i!=a;++i) { x=end1[i]-be1[i]; if(x>y) y=x; b=0; //d保存某个数轴段跟前一个数轴段的距离,若是此数轴段被包含在另一个数轴段里,则跳出循环 int d=0; //算最大不挤奶的时间时,就要考虑到重复的情况了 for(int j=0;j!=a;++j) { if(be1[i]==be1[j]&&end1[i]==end1[j]) continue; else if(end1[j]>=be1[i]&&end1[j]<=end1[i]&&be1[j]<=be1[i]) { d=0; break; } else if(be1[j]>=be1[i]&&end1[j]<=be1[i]) { d=0; break; } else if(be1[j]<=be1[i]&&end1[j]>=end1[i]) { d=0; break; } else if(be1[j]<=end1[i]&&be1[j]>=be1[i]&&end1[j]>=end1[i]) { d=0; break; } if(be1[i]>end1[j]) b=be1[i]-end1[j]; if(end1[i]<be1[j]) continue; if(d==0&&b!=0) d=b; //相当于D的初始化 if(b<d) d=b; } if(d>c) c=d; } fout<<y<<' '<<c<<endl; return 0; }

把时间看过一个数轴,然后合并重复的时间,但是没有排序,没有官方答案的简洁!

 

官方答案:

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #define MAXMILKING 5000 //用结构体会比较方便,要学会用结构体 typedef struct Milking Milking; struct Milking { int begin; int end; }; Milking milking[MAXMILKING]; int nmilking; //qsort算法的函数,要学会用qsort算法,很方便的排序 int milkcmp(const void *va, const void *vb) { Milking *a, *b; a = (Milking*)va; b = (Milking*)vb; if(a->begin > b->begin) return 1; if(a->begin < b->begin) return -1; return 0; } void main(void) { FILE *fin, *fout; int i, j, t, tmilk, tnomilk; Milking cur; fin = fopen("milk2.in", "r"); fout = fopen("milk2.out", "w"); assert(fin != NULL && fout != NULL); /* read input, sort */ fscanf(fin, "%d", &nmilking); for(i=0; i<nmilking; i++) fscanf(fin, "%d %d", &milking[i].begin, &milking[i].end); qsort(milking, nmilking, sizeof(Milking), milkcmp); /* walk over list, looking for long periods of time */ /* tmilk = longest milking time */ /* tnomilk = longest non-milking time */ /* cur = current span of milking time being considered */ tmilk = 0; tnomilk = 0; cur = milking[0]; //一边计算,一边合并重复的时间,很值得学习 for(i=1; i<nmilking; i++) { if(milking[i].begin > cur.end) { /* a down time */ t = milking[i].begin - cur.end; if(t > tnomilk) tnomilk = t; t = cur.end - cur.begin; if(t > tmilk) tmilk = t; cur = milking[i]; } else { if(milking[i].end > cur.end) cur.end = milking[i].end; } } /* check final milking period */ t = cur.end - cur.begin; if(t > tmilk) tmilk = t; fprintf(fout, "%d %d/n", tmilk, tnomilk); exit(0); }

 

排序算法很值得学习!

 

We read the list of times, sort it by start time, and then walk over the list once, merging overlapping times. Then we walk the list watching for long milking periods and long non-milking periods.

An alternate approach would be to just keep an array of size a million and mark off times. On a nice fast processor, that's probably fast enough, but our above algorithm will work even on slow processors, and it's not much harder to write.