zoukankan      html  css  js  c++  java
  • HDU 2686 Matrix

    Matrix

    Time Limit: 1000ms
    Memory Limit: 32768KB
    This problem will be judged on HDU. Original ID: 2686
    64-bit integer IO format: %I64d      Java class name: Main
    Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.
    Every time yifenfei should to do is that choose a detour which frome the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix yifenfei choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And yifenfei can not pass the same area of the Matrix except the start and end. 
     

    Input

    The input contains multiple test cases.
    Each case first line given the integer n (2<n<30) 
    Than n lines,each line include n positive integers.(<100)
     

    Output

    For each test case output the maximal values yifenfei can get.
     

    Sample Input

    2
    10 3
    5 10
    3
    10 3 3
    2 5 3
    6 7 10
    5
    1 2 3 4 5
    2 3 4 5 6
    3 4 5 6 7
    4 5 6 7 8
    5 6 7 8 9

    Sample Output

    28
    46
    80

    Source

     
    解题:传说中的双进程dp
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 int dp[80][40][40],val[40][40],n;
     4 int main() {
     5     while(~scanf("%d",&n)) {
     6         memset(dp,0,sizeof dp);
     7         for(int i = 0; i < n; ++i)
     8             for(int j = 0; j < n; ++j)
     9                 scanf("%d",val[i] + j);
    10         for(int k = 1; k < 2*n - 2; ++k) {
    11             for(int i = 0; i < n; ++i)
    12                 for(int j = 0; j < n; ++j) {
    13                     if(i == j) continue;
    14                     dp[k][i][j] = max(max(dp[k-1][i][j],dp[k-1][i-1][j]),max(dp[k-1][i][j-1],dp[k-1][i-1][j-1]));
    15                     dp[k][i][j] += val[i][k - i] + val[j][k - j];
    16                 }
    17         }
    18         int t = 2*n-3;
    19         printf("%d
    ",max(dp[t][n-1][n-2],dp[t][n-2][n-1]) + val[0][0] + val[n-1][n-1]);
    20     }
    21     return 0;
    22 }
    View Code
  • 相关阅读:
    Windows出现BOOTBCD错误的解决办法
    解决真机调试时Eclipse DDMS上打不开/data目录的问题
    解决魅族MX5卸载debug-app不干净,导致安装、升级不成功的问题
    iOS10 相机相册等权限的使用、检测并引导用户开启权限
    python脚本后台执行
    CentOS 6.8 安装 Python3
    Virtualenv教程
    Linux路由表
    python websocket-client connection
    RESTful API 设计指南
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4925131.html
Copyright © 2011-2022 走看看