Islands Travel——SPFA求最短路

又是一道微软的笔试题，看来最短路径很重要啊！

时间限制:10000ms   单点时限:1000ms   内存限制:256MB

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-Xj|, |Yi-Yj|} (|a| denotes the absolute value of a. min{a, b} denotes the smaller value between a and b) gold coins. You want to know what is the minimum cost to travel from the 1st island to the n-th island.

输入

Line 1: an integer N.

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

For 40% data, N<=1000，0<=Xi,Yi<=100000.

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

输出

Output the minimum cost.

样例输入

32 21 77 6

样例输出

2

掌握SPFA，这道题就很简单了，学习链接：

SPFA算法解析

make_pair教程

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<vector>#include<queue>using namespace std;const int maxn = 200000;//数组要开大些，防止越界int n , dis[maxn] , isQueue[maxn];vector < pair < int , int > > graph[maxn];//注意后面两个尖括号之前有空格！class nodd{public:    int x , y , num;    //重载"<"符号    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 ) { //当两个点的x值相等时            int a = A[nex].num , b = A[i].num; //排序后点的序号改变，不能直接用i表示，建立新的变量获取该点原来的序号            graph[a].push_back( make_pair( b , 0 ) );//建立一条从a到b的边，距离为0            graph[b].push_back( make_pair( a , 0 ) );//因为是无向图，同样建立一条从b到a的边，距离为0            nex ++; //把所有x相等的点连起来        }        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; //因为while中进行nex++操作，所以i要从最后一个x与前面一个点x值相同的点开始    }}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(); //将x之间的差作为距离    for ( int i = 0; i < n; i ++ )        swap( A[i].x , A[i].y );    build(); //将y之间的差作为距离    queue < int > q;    for ( int i = 0; i < n; i ++ )        dis[i] = 2000000000;    dis[0] = 0;    q.push( 0 ); //第一个点入队    isQueue[0] = 1;    while ( !q.empty() ) {        int now = q.front() , nex;        q.pop(); //当前点出队        isQueue[now] = 0;        for ( int i = graph[now].size()-1; i >= 0; i -- ) { //遍历跟当前点有边的所有点            nex = graph[now][i].first; //获得与当前点相连的点的编号            if ( dis[now] + graph[now][i].second < dis[nex] ) { //当前点到起点的距离+当前点到下一个点的距离<下一个点到起点的距离                dis[nex] = dis[now] + graph[now][i].second;                if ( isQueue[nex] == 0 ) { //如果下一个点不在队列中，入队；否则让它留在队中供其他点做判断                    isQueue[nex] = 1 ;                    q.push( nex );                }            }        }    }    cout << dis[n-1] << endl;    return 0;}