hdoj 2391 Filthy Rich (动态规划)
2015-03-22 17:38
232 查看
Problem I
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 84 Accepted Submission(s) : 28
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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 throughoutthe 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.此处周赛没看到直接PE了一次
Sample Input
1 3 4 1 10 8 8 0 0 1 8 0 27 0 4
Sample Output
Scenario #1: 42
/*
典型的动态规划
*/
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
int map[1001][1001];
int dp[1001][1001];
int Max(int a,int b,int c)
{
return max(a,max(b,c));
}
int main()
{
int t,ca=1;
scanf("%d",&t);
while(t--)
{
int r,c,i,j;
scanf("%d%d",&r,&c);
for(i=1;i<=r;i++)
for(j=1;j<=c;j++)
scanf("%d",&map[i][j]);
memset(dp,0,sizeof(dp));
for(i=1;i<=r;i++)
for(j=1;j<=c;j++)
dp[i][j]+=Max(dp[i-1][j],dp[i-1][j-1],dp[i][j-1])+map[i][j];
printf("Scenario #%d:\n%d\n\n",ca++,dp[r][c]);//注意格式
}
return 0;
}
相关文章推荐
- (精)hdoj 2391(动态规划)
- hdoj 2391 Filthy Rich(dp)不是搜素
- HDOJ 2391 Filthy Rich
- hdoj-2391 Filthy Rich
- hdoj 2391 Filthy Rich 【DP】
- HDOJ 2391 Filthy Rich
- HDU 2391 Filthy Rich(动态规划)
- hdoj 2391 Filthy Rich
- 【动态规划】HDOJ 1257 最少拦截系统
- hdu 2391 Filthy Rich
- hdoj-1159-Common Subsequence【动态规划求最长公共子序列】
- HDOJ Hatsune Miku 5074【2014鞍山区域赛E题-动态规划】
- hdoj problem 2845 Beans (动态规划)
- 杭电OJ(HDOJ)1231题:最大连续子序列(动态规划)
- HDOJ 2571 命运(动态规划)
- HDOJ 1024 Max Sum Plus Plus -- 动态规划
- hdoj 1003 Max Sum(动态规划)
- HDOJ3336 Count the string 【KMP前缀数组】+【动态规划】
- HDOJ2059龟兔赛跑--动态规划问题
- HDOJ 动态规划总结