又是一道微軟的筆試題，看來最短路徑很重要啊！

描述

There are N islands on a planet whose coordinates are (X1, Y1), (X2, Y2), (X3, Y3) ..., (XN, YN). You starts at the 1st island (X1, Y1) and your destination is the n-th island (XN, YN). Travelling between i-th and j-th islands will cost you min{|Xi

輸入

Line 1: an integer N.

Line 2~N+1: each line contains two integers Xi and Yi.

For 40% data, N<=1000，0<=Xi

For 100% data, N<=100000，0<=Xi,Yi<=1000000000.

輸出

Output the minimum cost.

3
2 2
1 7
7 6

2

掌握SPFA，這道題就很簡單了，學習連結：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 200000;//陣列要開大些，防止越界
int n , dis[maxn] , isQueue[maxn];
vector < pair < int , int > > graph[maxn];//注意後面兩個尖括號之前有空格！

class nodd
{
public:
    int x , y , num;
    //過載"<"符號
    bool operator < (const nodd  A) const {
        return x < A.x;
    }
} A[maxn];

void build()
{
    sort( A , A + n );
    for ( int i = 0; i < n; i ++ ) {
        int nex = i + 1;
        while ( nex < n && A[nex].x == A[i].x ) { //當兩個點的x值相等時
            int a = A[nex].num , b = A[i].num; //排序後點的序號改變，不能直接用i表示，建立新的變數獲取該點原來的序號
            graph[a].push_back( make_pair( b , 0 ) );//建立一條從a到b的邊，距離為0
            graph[b].push_back( make_pair( a , 0 ) );//因為是無向圖，同樣建立一條從b到a的邊，距離為0
            nex ++; //把所有x相等的點連起來
        }
        if ( i > 0 ) {
            int a = A[i].num , b = A[i-1].num;
            graph[a].push_back( make_pair( b , A[i].x - A[i-1].x ) );
            graph[b].push_back( make_pair( a , A[i].x - A[i-1].x ) );
        }
        i = nex - 1; //因為while中進行nex++操作，所以i要從最後一個x與前面一個點x值相同的點開始
    }
}

int main()
{
    scanf("%d",&n);
    for ( int i = 0; i < n; i ++ ) {
        scanf("%d%d",&A[i].x,&A[i].y);
        A[i].num = i;
    }
    build(); //將x之間的差作為距離
    for ( int i = 0; i < n; i ++ )
        swap( A[i].x , A[i].y );
    build(); //將y之間的差作為距離

    queue < int > q;
    for ( int i = 0; i < n; i ++ )
        dis[i] = 2000000000;
    dis[0] = 0;
    q.push( 0 ); //第一個點入隊
    isQueue[0] = 1;
    while ( !q.empty() ) {
        int now = q.front() , nex;
        q.pop(); //當前點出隊
        isQueue[now] = 0;
        for ( int i = graph[now].size()-1; i >= 0; i -- ) { //遍歷跟當前點有邊的所有點
            nex = graph[now][i].first; //獲得與當前點相連的點的編號
            if ( dis[now] + graph[now][i].second < dis[nex] ) { //當前點到起點的距離+當前點到下一個點的距離<下一個點到起點的距離
                dis[nex] = dis[now] + graph[now][i].second;
                if ( isQueue[nex] == 0 ) { //如果下一個點不在佇列中，入隊；否則讓它留在隊中供其他點做判斷
                    isQueue[nex] = 1 ;
                    q.push( nex );
                }
            }
        }
    }
    cout << dis[n-1] << endl;
    return 0;
}