zoukankan      html  css  js  c++  java
  • Metro-Ural119递推

    Time limit: 0.5 second Memory limit: 64 MB

    Many of SKB Kontur programmers like to get to work by Metro because the main office is situated quite close the station Uralmash. So, since a sedentary life requires active exercises off-duty, many of the staff — Nikifor among them — walk from their homes to Metro stations on foot.
    这里写图片描述
    Problem illustration
    Nikifor lives in a part of our city where streets form a grid of residential quarters. All the quarters are squares with side 100 meters. A Metro entrance is situated at one of the crossroads. Nikifor starts his way from another crossroad which is south and west of the Metro entrance. Naturally, Nikifor, starting from his home, walks along the streets leading either to the north or to the east. On his way he may cross some quarters diagonally from their south-western corners to the north-eastern ones. Thus, some of the routes are shorter than others. Nikifor wonders, how long is the shortest route.
    You are to write a program that will calculate the length of the shortest route from the south-western corner of the grid to the north-eastern one.

    Input

    There are two integers in the first line: N and M (0 < N,M ≤ 1000) — west-east and south-north sizes of the grid. Nikifor starts his way from a crossroad which is situated south-west of the quarter with coordinates (1, 1). A Metro station is situated north-east of the quarter with coordinates (N, M). The second input line contains a number K (0 ≤ K ≤ 100) which is a number of quarters that can be crossed diagonally. Then K lines with pairs of numbers separated with a space follow — these are the coordinates of those quarters.
    Output
    Your program is to output a length of the shortest route from Nikifor’s home to the Metro station in meters, rounded to the integer amount of meters.

    Sample

    input
    3 2
    3
    1 1
    3 2
    1 2
    output
    383

    Problem Author:

    Leonid Volkov

    Problem Source:

    USU Open Collegiate Programming Contest October’2001 Junior Session

    对于图中的某一点ij,则Dp[i][j]表示从(0,0)到(i,j)的最短距离。所以对于(i,j)可以到达他的点为(i-1,j)(i,j-1)和(i-1,j-1)(如果可以),所以可以从下到上,从左到右推过去。

    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <string>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    
    const int Max = 1010;
    
    const double len = 100*sqrt(2);
    
    double Dp[Max][Max];
    
    bool Map[Max][Max];
    
    int n,m;
    
    void Init()
    {
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=m;j++)
            {
                Map[i][j]=false;
    
                Dp[i][j] = INF;
            }
        }
    }
    
    bool Judge(int x,int y)
    {
        if(x<=n&&y<=m)
        {
            return true;
        }
        return false;
    }
    
    int main()
    {
        int num;
    
        int u,v;
    
        while(~scanf("%d %d",&m,&n))
        {
            Init();
    
            scanf("%d",&num);
    
            while(num--)
            {
                scanf("%d %d",&u,&v);
    
                Map[v-1][u-1] = true;
    
            }
            Dp[0][0]=0;
            for(int i=0;i<=n;i++)
            {
                for(int j=0;j<=m;j++)
                {
                    if(Judge(i+1,j))
                    {
                        Dp[i+1][j]=  min(Dp[i+1][j],Dp[i][j]+100);
                    }
                    if(Judge(i,j+1))
                    {
                        Dp[i][j+1] = min(Dp[i][j+1],Dp[i][j]+100);
                    }
                    if(Map[i][j]&&Judge(i+1,j+1))
                    {
                        Dp[i+1][j+1]=min(Dp[i+1][j+1],Dp[i][j]+len);
                    }
                }
            }
    
            printf("%.0f
    ",Dp[n][m]);
        }
        return 0;
    }
    
  • 相关阅读:
    Angle Beats Gym
    MUV LUV UNLIMITED Gym
    Balanced Diet Gym
    数位dp HDU
    数位dp CodeForces
    数位dp HDU
    有依赖的背包 洛谷P1064 金明的预算方案 (不是分组背包)
    多重背包+二进制拆分 POJ1014
    单调队列优化dp 入门 洛谷P2627 修剪草坪
    01背包 + 排序 (记忆化搜索) 骄傲的商人(HDU
  • 原文地址:https://www.cnblogs.com/juechen/p/5255867.html
Copyright © 2011-2022 走看看