首页 > 代码库 > POJ 3176 Cow Bowling 保龄球 数塔问题 DP
POJ 3176 Cow Bowling 保龄球 数塔问题 DP
题目链接：POJ 3176 Cow Bowling
Cow Bowling
Description The cows don‘t use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: 7
3 8
8 1 0
2 7 4 4
4 5 2 6 5Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow‘s score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame. Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable. Input Line 1: A single integer, N Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle. Output Line 1: The largest sum achievable using the traversal rules Sample Input 5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 Sample Output 30 Hint Explanation of the sample: 7
*
3 8
*
8 1 0
*
2 7 4 4
*
4 5 2 6 5The highest score is achievable by traversing the cows as shown above.Source USACO 2005 December Bronze |
题意：
从数塔的第一层走到最底层，但只能沿对角线走，求路径上的数的和的最大值。
分析：
从最底层向上考虑，路径上的和的大小取决于直接取决于下面两个数的大小。因而采用自顶向上方法，逐步向上走，走到最顶层，每次都选择最大的和，这样最后的结果就保存在了最顶层。
状态转移方程：dp[i][j] = max(dp[i+1][j], dp[i+1][j+1])+A[i][j];
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX_N = 400;
int dp[MAX_N][MAX_N];
int N, A[MAX_N][MAX_N];
void solve() {
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= N; i++) dp[N][i] = A[N][i];
for(int i = N-1; i >= 1; i--)
for(int j = 1; j <= i; j++)
dp[i][j] = A[i][j]+max(dp[i+1][j], dp[i+1][j+1]);
printf("%d\n", dp[1][1]);
}
int main() {
while(~scanf("%d", &N)) {
for(int i = 1; i <= N; i++)
for(int j = 1; j <= i; j++)
scanf("%d", &A[i][j]);
solve();
}
return 0;
}POJ 3176 Cow Bowling 保龄球 数塔问题 DP
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。