Codeforces Round #355 (Div. 2) D. Vanya and Treasure 分治暴力

D. Vanya and Treasure

题目连接：

http://www.codeforces.com/contest/677/problem/D

Description

Vanya is in the palace that can be represented as a grid n × m. Each room contains a single chest, an the room located in the i-th row and j-th columns contains the chest of type aij. Each chest of type x ≤ p - 1 contains a key that can open any chest of type x + 1, and all chests of type 1 are not locked. There is exactly one chest of type p and it contains a treasure.

Vanya starts in cell (1, 1) (top left corner). What is the minimum total distance Vanya has to walk in order to get the treasure? Consider the distance between cell (r1, c1) (the cell in the row r1 and column c1) and (r2, c2) is equal to |r1 - r2| + |c1 - c2|.

Input

The first line of the input contains three integers n, m and p (1 ≤ n, m ≤ 300, 1 ≤ p ≤ n·m) — the number of rows and columns in the table representing the palace and the number of different types of the chests, respectively.

Each of the following n lines contains m integers aij (1 ≤ aij ≤ p) — the types of the chests in corresponding rooms. It's guaranteed that for each x from 1 to p there is at least one chest of this type (that is, there exists a pair of r and c, such that arc = x). Also, it's guaranteed that there is exactly one chest of type p.

The only line of the input contains a single word s (1 ≤ |s| ≤ 100 000), consisting of digits, lowercase and uppercase English letters, characters '-' and '_'.

Output

Print one integer — the minimum possible total distance Vanya has to walk in order to get the treasure from the chest of type p.

Sample Input

3 4 3

2 1 1 1

1 1 1 1

2 1 1 3

Sample Output

5

Hint

题意

给你一个n*m的方格，你需要把权值为p的那个门打开，前提是你必须打开过权值为p-1的门……

然后问你打开权值为p的门，你最少走的距离是多少，一开始你在(1,1)

题解：

n*mlog的就扫描线莽一波

n^3的话，就分治一下就好了，如果当前权值的数量小于sqrt(nm)的话，我就暴力更新，否则我就在全图进行一次bfs

这样均摊下来，复杂度是nmsqrt(n)sqrt(m)的

所以直接暴力吧

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 305;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int a[maxn][maxn],dp[maxn][maxn],n,m,p;
int dis[maxn][maxn],inq[maxn][maxn];
vector<pair<int,int> >V[maxn*maxn];
int main()
{
    memset(dp,127,sizeof(dp));
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&a[i][j]);
            if(a[i][j]==1)dp[i][j]=i-1+j-1;
            V[a[i][j]].push_back(make_pair(i,j));
        }
    }
    int level = sqrt(n*m);
    int idx=1;
    queue<pair<int,int> >Q;
    for(int i=1;i<p;i++)
    {
        int sz = V[i].size();
        if(sz<=level){
            for(auto v:V[i+1])
                for(auto u:V[i])
                    dp[v.first][v.second]=min(dp[v.first][v.second],dp[u.first][u.second]+abs(v.first-u.first)+abs(v.second-u.second));
        }
        else{
            idx++;
            for(auto v:V[i])
                Q.push(v);
            memset(dis,127,sizeof(dis));
            for(auto v:V[i])
                dis[v.first][v.second]=dp[v.first][v.second];
            while(!Q.empty())
            {
                pair<int,int>now=Q.front();
                Q.pop();
                inq[now.first][now.second]=0;
                for(int k=0;k<4;k++)
                {
                    pair<int,int>next=now;
                    next.first+=dx[k];
                    next.second+=dy[k];
                    if(next.first<1||next.first>n)continue;
                    if(next.second<1||next.second>m)continue;
                    if(dis[next.first][next.second]>dis[now.first][now.second]+1)
                    {
                        dis[next.first][next.second]=dis[now.first][now.second]+1;
                        if(inq[next.first][next.second]<idx){
                            inq[next.first][next.second]=idx;
                            Q.push(next);
                        }
                    }
                }
            }
            for(auto v:V[i+1])
                dp[v.first][v.second]=dis[v.first][v.second];
        }
    }
    cout<<dp[V[p][0].first][V[p][0].second]<<endl;
}