HDU 5492 Find a path

最新推荐文章于 2025-09-12 17:31:31 发布

weixin_30797199

最新推荐文章于 2025-09-12 17:31:31 发布

阅读量51

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/crackpotisback/p/4868265.html

解决一个迷宫问题，寻找从起点到终点路径中方差最小的路径。采用动态规划方法，通过记录每个点上的累计和与累计平方和来计算路径的方差。

Find a path

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 5492
64-bit integer IO format: %I64d Java class name: Main

Frog fell into a maze. This maze is a rectangle containing N rows and M columns. Each grid in this maze contains a number, which is called the magic value. Frog now stays at grid (1, 1), and he wants to go to grid (N, M). For each step, he can go to either the grid right to his current location or the grid below his location. Formally, he can move from grid (x, y) to (x + 1, y) or (x, y +1), if the grid he wants to go exists.
Frog is a perfectionist, so he'd like to find the most beautiful path. He defines the beauty of a path in the following way. Let’s denote the magic values along a path from (1, 1) to (n, m) as $A_1,A_2,…A_{N+M−1}$, and $A_{avg}$ is the average value of all $A_i$. The beauty of the path is (N+M–1) multiplies the variance of the values:$(N+M−1)\sum_{i=1}^{N+M−1}(A_i−A_{avg})^2$
In Frog's opinion, the smaller, the better. A path with smaller beauty value is more beautiful. He asks you to help him find the most beautiful path.

Input
The first line of input contains a number T indicating the number of test cases $(T\leq 50).$
Each test case starts with a line containing two integers N and M $(1\leq N,M\leq 30)$. Each of the next N lines contains M non-negative integers, indicating the magic values. The magic values are no greater than 30.

Output
For each test case, output a single line consisting of “Case #X: Y”. X is the test case number starting from 1. Y is the minimum beauty value.

Sample Input
1
2 2
1 2
3 4

Sample Output
Case #1: 14

Source
2015 ACM/ICPC Asia Regional Hefei Online

解题：动态规划，最小方差路

$dp[i][j][k]表示在(i,j)格子中\sum{A_i}为k的时候最小的\sum{A_i^2}$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int INF = 0x3f3f3f3f;
 4 const int maxn = 35;
 5 int mp[maxn][maxn],dp[maxn][maxn][1810];
 6 int main(){
 7     int kase,n,m,cs = 1;
 8     scanf("%d",&kase);
 9     while(kase--){
10         scanf("%d%d",&n,&m);
11         for(int i = 1; i <= n; ++i)
12             for(int j = 1; j <= m; ++j)
13                 scanf("%d",mp[i] + j);
14         memset(dp,0x3f,sizeof dp);
15         dp[1][0][0] = dp[0][1][0] = 0;
16         for(int i = 1; i <= n; ++i){
17             for(int j = 1; j <= m; ++j){
18                 for(int k = 0; k <= 1800; ++k){
19                     if(dp[i-1][j][k] != INF)
20                         dp[i][j][k + mp[i][j]] = min(dp[i][j][k + mp[i][j]],dp[i-1][j][k] + mp[i][j]*mp[i][j]);
21                     if(dp[i][j-1][k] != INF)
22                         dp[i][j][k + mp[i][j]] = min(dp[i][j][k + mp[i][j]],dp[i][j-1][k] + mp[i][j]*mp[i][j]);
23                 }
24             }
25         }
26         int ret = INF;
27         for(int i = 0; i <= 1800; ++i)
28             if(dp[n][m][i] < INF) ret = min(ret,(n + m - 1)*dp[n][m][i] - i*i);
29         printf("Case #%d: %d\n",cs++,ret);
30     }
31     return 0;
32 }