Codeforces 383B. Volcanoes 模拟

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本文介绍了一种在大型沙漠迷宫中寻找最短路径的方法，考虑到火山障碍的存在，需要找到从起点(1,1)到终点(n,n)的最快路径。通过记录每一层可以移动的区间来解决该问题。

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记录每一层可以移动到的区间....非常多的细节...

Iahub got lost in a very big desert. The desert can be represented as a n × n square matrix, where each cell is a zone of the desert. The cell (i, j) represents the cell at row i and column j (1 ≤ i, j ≤ n). Iahub can go from one cell (i, j) only down or right, that is to cells (i + 1, j)or (i, j + 1).

Also, there are m cells that are occupied by volcanoes, which Iahub cannot enter.

Iahub is initially at cell (1, 1) and he needs to travel to cell (n, n). Knowing that Iahub needs 1 second to travel from one cell to another, find the minimum time in which he can arrive in cell (n, n).

Input

The first line contains two integers n (1 ≤ n ≤ 109) and m (1 ≤ m ≤ 105). Each of the next m lines contains a pair of integers, x and y (1 ≤ x, y ≤ n), representing the coordinates of the volcanoes.

Consider matrix rows are numbered from 1 to n from top to bottom, and matrix columns are numbered from 1 to n from left to right. There is no volcano in cell (1, 1). No two volcanoes occupy the same location.

Output

Print one integer, the minimum time in which Iahub can arrive at cell (n, n). If no solution exists (there is no path to the final cell), print -1.

Sample test(s)

input
4 2
1 3
1 4

output
6

input
7 8
1 6
2 6
3 5
3 6
4 3
5 1
5 2
5 3

output
12

input
2 2
1 2
2 1

output
-1

Note

Consider the first sample. A possible road is: (1, 1) → (1, 2) → (2, 2) → (2, 3) → (3, 3) → (3, 4) → (4, 4).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

int n,m;

typedef pair<int,int> pII;

vector<pII> vi;
vector<pII> duan[2],temp;

int hy[320000],ny=0;
int MX=-1;

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        vi.push_back(make_pair(a,b));
        if(a-1>=1) hy[ny++]=a-1;
        hy[ny++]=a;
        if(a+1<=n) hy[ny++]=a+1;
    }
    hy[ny++]=n; hy[ny++]=1;
    sort(hy,hy+ny);
    ny=unique(hy,hy+ny)-hy;
    sort(vi.begin(),vi.end());
    int p=0;
    int pre=0,now=1;

    int left=-1;
    if(vi[p].first==1)
    {
        for(;p<vi.size();p++)
        {
            if(vi[p].first==1)
            {
                if(left==-1) left=vi[p].second-1;
                else left=min(left,vi[p].second-1);
            }
            else
            {
                p--; break;
            }
        }
        if(left<1)
        {
            puts("-1");
            return 0;
        }
        else duan[0].push_back(make_pair(1,left));
    }
    else duan[0].push_back(make_pair(1,n));
    p=0;
    for(int yy=0;yy<ny;yy++)
    {
        temp.clear();
        int one=1,two;
        while(vi[p].first==hy[yy])
        {
            two=vi[p].second-1;
            if(two>=one)
            {
                temp.push_back(make_pair(one,two));
            }
            one=two+2;
            p++;
        }
        if(one<=n) temp.push_back(make_pair(one,n));
        duan[now].clear();
        int id=0;
        for(int i=0,sz=duan[pre].size(),ts=temp.size();i<sz&&id<ts;i++)
        {
            int from=duan[pre][i].first;
            int to=duan[pre][i].second;
            if(to<temp[id].first) continue;
            while(id+1<ts&&temp[id].second<from)
                id++;
            if(to>=temp[id].first&&from<=temp[id].second)
            {
                duan[now].push_back( make_pair( max(from,temp[id].first)
                                    , temp[id].second) );
                if(hy[yy]==n)
                {
                    MX=max(MX,temp[id].second);
                }
                id++; i--;
            }
        }
        swap(now,pre);
    }
    if(MX>=n) printf("%d\n",2*n-2);
    else printf("-1\n");
    return 0;
}