HDoj2391-Filthy Rich-DP
来源:互联网 发布:js获取audio时长 编辑:程序博客网 时间:2024/05/22 05:26
Filthy Rich
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2097 Accepted Submission(s): 944
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.
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.
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#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<queue>#include<stack>using namespace std;int dp[1010][1010];int main(){int T,t;cin>>T;for(t=1;t<=T;t++){int m,n;cin>>n>>m;memset(dp,0,sizeof(dp));int i,j;for(i=1;i<=n;i++)for(j=1;j<=m;j++)cin>>dp[i][j];for(i=1;i<=n;i++) //起始位置从dp[1][1]开始for(j=1;j<=m;j++)dp[i][j]+=max(dp[i-1][j-1],max(dp[i-1][j],dp[i][j-1]));cout<<"Scenario #"<<t<<":"<<endl;cout<<dp[n][m]<<endl<<endl;}return 0;}
0 0
- HDoj2391-Filthy Rich-DP
- Filthy Rich+DP
- HDU2391 Filthy Rich 【DP】
- HDU2391 Filthy Rich(DP)
- hdoj 2391 Filthy Rich 【DP】
- hdoj 2391 Filthy Rich 【DP】
- HDU2391 Filthy Rich【数塔DP】
- HDOJ 2391 Filthy Rich (简单DP)
- HDU 2391 Filthy Rich (滚动dp)
- HDU 2391 Filthy Rich (简单DP)
- HDU 2391 Filthy Rich(dp)
- HDU 2391 Filthy Rich(dp)
- hdu 2391 Filthy Rich(DP)
- Filthy Rich
- Filthy Rich
- HDOJ 2391 Filthy Rich dp动态规划.....so easy......
- HDOJ 题目2391Filthy Rich(水dp)
- hdoj 2391 Filthy Rich(dp)不是搜素
- android中单选列表对话框-选择个人特长
- 再看一遍,只想说,第一次真的没认真看,好多一点印象也没有,却实实在在写在哪里,
- Android摇一摇动作源码详解
- 100 hours anti boiled water film faced plywood
- 解决tableview出现多余行
- HDoj2391-Filthy Rich-DP
- UVALive - 3644 X-Plosives 并查集
- 欧拉函数
- 萧夜迷离
- php 让浏览器格式化显示XML
- 链表的基本操作
- iOS UIFont 字体大全
- Shell常用命令整理
- 一道超经典易错题