1046 Shortest Distance (20 分)

标签：really   short   void   print   lse   时间   复杂度   result   guarantee   

1046 Shortest Distance (20 分)

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3,105]), followed by N integer distances D1 D2 ? DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (≤104), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 107.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

一、改进解答

#include<stdio.h>
#include<stdlib.h>
#define MaxN 100001

int GettingData(int *D);
int main()
{
	int sum,dis,antidis;
    int m,i;
    int s,d,tmp;
	int distance[MaxN];
	sum=GettingData(distance);
	scanf("%d",&m);
	for(i=0;i<m;i++){
		scanf(" %d %d",&s,&d);
		if(s>d){tmp=s;s=d;d=tmp;}
 		dis=distance[d]-distance[s];
 		antidis=sum-dis;
 		if(dis<antidis) printf("%d\n",dis);
		else printf("%d\n",antidis);
	}
	return 0;
 } 
  
int GettingData(int *D)
 {
 	int i,t,sum=0;
 	scanf("%d",&D[0]);
 	for(i=1;i<=D[0];i++){
 		scanf("%d",&t);
 		D[i]=sum;
 		sum+=t;
	 }	
 	return sum;
 }

二、不太聪明的解答：

每组测试数据都要遍历一遍数组，时间复杂度接近O(N)，超时了，若对数据进行处理，时间复杂度可降低为O(1)。

#include<stdio.h>
#include<stdlib.h>
#define MaxN 100001
#define MaxM 20001

void GettingData(int *D,int *T);
void ShortestDistance(int *D,int *T);
int main()
{
	int distance[MaxN],test[MaxM];
	GettingData(distance,test);
	ShortestDistance(distance,test);
	return 0;
 } 
 
 void ShortestDistance(int *D,int *T)
 {
 	int i=1,j,tmp;
 	int s,d;
 	int dis,antidis;
 	while(i<2*T[0]){
 		dis=0,antidis=0;
 		s=T[i++];d=T[i++];
 		if(s>d){tmp=s;s=d;d=tmp;}
 		if(s<0||d>D[0]) break;
 		for(j=s;j!=d;j=j%D[0]+1)
 			dis+=D[j];
		for(j=d;j!=s;j=j%D[0]+1)
			antidis+=D[j];
		if(dis<antidis) printf("%d\n",dis);
		else printf("%d\n",antidis);
	 }
 }
 
 void GettingData(int *D,int *T)
 {
 	int i;
 	scanf("%d",&D[0]);
 	for(i=1;i<=D[0];i++)
 		scanf("%d",&D[i]);
 	scanf("%d",&T[0]);
 	for(i=1;i<=2*T[0];i++)
 		scanf(" %d",&T[i]);
 	return;
 }

1046 Shortest Distance (20 分)

标签：really   short   void   print   lse   时间   复杂度   result   guarantee   

原文地址：https://www.cnblogs.com/present1440/p/14964899.html