HDU 2391 Filthy Rich (滚动dp)

Filthy Rich

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2651    Accepted Submission(s): 1176

Problem Description

They say that in Phrygia, the streets are paved with gold. You’re currently on vacation in Phrygia, and to your astonishment you discover that this is to be taken literally: small heaps of gold are distributed throughout the city. On a certain day, the Phrygians even allow all the tourists to collect as much gold as they can in a limited rectangular area. As it happens, this day is tomorrow, and you decide to become filthy rich on this day. All the other tourists decided the same however, so it’s going to get crowded. Thus, you only have one chance to cross the field. What is the best way to do so?

Given a rectangular map and amounts of gold on every field, determine the maximum amount of gold you can collect when starting in the upper left corner of the map and moving to the adjacent field in the east, south, or south-east in each step, until you end up in the lower right corner.

 

Input

The input starts with a line containing a single integer, the number of test cases.
Each test case starts with a line, containing the two integers r and c, separated by a space (1 <= r, c <= 1000). This line is followed by r rows, each containing c many integers, separated by a space. These integers tell you how much gold is on each field. The amount of gold never negative.
The maximum amount of gold will always fit in an int.

 

Output

For each test case, write a line containing “Scenario #i:”, where i is the number of the test case, followed by a line containing the maximum amount of gold you can collect in this test case. Finish each test case with an empty line.

 

Sample Input

13 41 10 8 80 0 1 80 27 0 4

 

Sample Output

Scenario #1:42

 

Source

HDU 2008-10 Public Contest

 

大体题意：

给你一个 r行c列的宝藏地图，每一个位置都有一定数量的宝藏。

你现在从左上角开始走一直走到右下角，问最大宝藏量是多少！

一个个人赛的题目，dfs 和bfs 不知怎么弄得 怎么搞都是爆内存~~

赛后才知道是dp，这么容易爆内存，弄个滚动dp就好了~

dp[i] [j] 表示i行j列位置最大宝藏数量。

处理：

第一行和第一列是特殊情况，第一行只能来自左方，第一列只能来自上方！

而其余的就可以来自三个方向了！

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1000 + 10;int dp[2][maxn];int mp[maxn][maxn];int main(){int T,cnt=0;scanf("%d",&T);while(T--){memset(dp,0,sizeof(dp));int r,c;scanf("%d%d",&r,&c);for (int i = 1; i <= r; ++i)for (int j = 1; j <= c; ++j)scanf("%d",&mp[i][j]);for (int i = 1; i <= c; ++i)dp[1][i] = mp[1][i] + dp[1][i-1];for (int i = 2; i <= r; ++i){for (int j = 1; j <= c; ++j){if (j == 1)dp[i&1][j] = dp[(i-1)&1][j] + mp[i][j];else dp[i&1][j] = mp[i][j] + max(dp[(i-1)&1][j],max(dp[i&1][j-1],dp[(i-1)&1][j-1]));}}printf("Scenario #%d:\n",++cnt);printf("%d\n",dp[r&1][c]);printf("\n");}return 0;}