hiho #1138 : Islands Travel 【最短路】

#1138 : Islands Travel

Time Limit:10000ms
Case Time Limit:1000ms
Memory Limit:256MB

Description

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.

Input

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.

Output

Output the minimum cost.

Sample Input

3
2 2
1 7
7 6

Sample Output

2

题意：给你n个岛屿的坐标，i到j的花费为min(abs(x[i]-x[j]), abs(y[i]-y[j]))。问你1-n的最小花费。

思路：n比较大，不能直接搞。需要把图简化，先按x坐标是否相等划分集合，集合内部的点边权为0，两个邻近的集合建边，同理搞下y。其它集合不需要建边，因为到达第i个集合的最优路径一定会经过其邻近集合。画下图就很清楚了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (2000000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
struct Edge{
int to, val, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int dist[MAXN]; bool vis[MAXN];
void init(){CLR(head, -1); edgenum = 0;}
void addEdge(int u, int v, int w){
Edge E = {v, w, head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
int n;
int Spfa()
{
CLR(dist, INF); CLR(vis, false);
queue<int> Q; Q.push(1); vis[1] = true; dist[1] = 0;
while(!Q.empty())
{
int u = Q.front(); Q.pop(); vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dist[v] > dist[u] + edge[i].val)
{
dist[v] = dist[u] + edge[i].val;
if(!vis[v])
{
vis[v] = true;
Q.push(v);
}
}
}
}
return dist
;
}
struct Node{
int x, y, id;
};
Node In[MAXN];
bool cmp1(Node a, Node b){
if(a.x != b.x) return a.x < b.x;
return a.y < b.y;
}
bool cmp2(Node a, Node b){
if(a.y != b.y) return a.y < b.y;
return a.x < b.x;
}
int main()
{
while(Ri(n) != EOF)
{
for(int i = 0; i < n; i++) Ri(In[i].x), Ri(In[i].y), In[i].id = i+1;
init();
sort(In, In+n, cmp1); int i = 0, j;
while(i < n)
{
j = i+1;
while(j < n && In[i].x == In[j].x)
{
addEdge(In[i].id, In[j].id, 0);
addEdge(In[j].id, In[i].id, 0);
j++;
}
if(j < n)
{
addEdge(In[i].id, In[j].id, In[j].x - In[i].x);
addEdge(In[j].id, In[i].id, In[j].x - In[i].x);
}
i = j;
}
i = 0; sort(In, In+n, cmp2);
while(i < n)
{
j = i+1;
while(j < n && In[i].y == In[j].y)
{
addEdge(In[i].id, In[j].id, 0);
addEdge(In[j].id, In[i].id, 0);
j++;
}
if(j < n)
{
addEdge(In[i].id, In[j].id, In[j].y - In[i].y);
addEdge(In[j].id, In[i].id, In[j].y - In[i].y);
}
i = j;
}
Pi(Spfa());
}
return 0;
}