PAT A1046 Shortest Distance A1paca 11月20日

## 题目

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 D i 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``````

## 代码

### 错误版

``````#include<stdio.h>

const int MAXN = 10005;

int main(void){
int dis[MAXN];
int N, M;
int tot = 0;
int a, b;
int st;

scanf("%d", &N);
for(int i = 1; i<=N; i++){
scanf("%d", &dis[i]);
tot += dis[i];
}

scanf("%d", &M);
for(int i=0; i<M; i++){
scanf("%d%d", &a, &b);
if(a > b){
int x = a;
a = b;
b = x;
}
st = 0;
while(a != b){
st +=dis[a++];
}
if(st<=tot/2)
printf("%d\n", st);
else
printf("%d\n", tot-st);
}

return 0;
}``````

orz

### 正确代码

``````#include<stdio.h>

const int MAXN = 100005;

int main(void){
int dis[MAXN], toa[MAXN];
int N, M;
int tot = 0;
int a, b;
int st;

scanf("%d", &N);
toa = 0;
for(int i = 1; i<=N; i++){
scanf("%d", &dis[i]);
tot += dis[i];
toa[i+1] = tot;
}

scanf("%d", &M);
for(int i=0; i<M; i++){
scanf("%d%d", &a, &b);
if(a > b){
int x = a;
a = b;
b = x;
}
st = toa[b]- toa[a];
if(st<=tot/2)
printf("%d\n", st);
else
printf("%d\n", tot-st);
}

return 0;
}``````

so~ 总结：原程序每次查询都要累加，在极端情况下有 10^5 次操作，而一共有 10^4 次查询，所以就是 10^9 次操作，这是无法在 100ms 内完成的。而第二个程序加了 toa 数组来记录每个 exit 到第一个的距离，这样对每个查询只要计算`toa[a]-toa[b]`，按术语来说是查询复杂度为 O(1)，大大减少了查询时间。
bingo~~