【动态规划】POJ-3176

一、题目

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   5

Then 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   5

The highest score is achievable by traversing the cows as shown above.

二、思路&心得

​简单的动态规划问题，从底向上依次扫描就行了，可以用直接用DP，也可以用记忆化搜索。

三、代码

1.DP：

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAX_N = 355;

int N;
int dp[MAX_N][MAX_N];

void solve() {
	for (int i = 0; i < N; i ++) {
		for (int j = 0; j <= i; j ++) {
			scanf("%d", &dp[i][j]);			
		}
	}
	for (int i = N - 2; i >=0; i --) {
		for (int j = 0; j <= i; j ++) {
			dp[i][j] += max(dp[i + 1][j], dp[i + 1][j + 1]);
		}
	}
	printf("%d\n", dp[0][0]);
}

int main() {
	while (~scanf("%d", &N)) {
		solve();
	}
	return 0;
}

2.记忆化搜索：

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAX_N = 355;

int N;
int dp[MAX_N][MAX_N];
int visit[MAX_N][MAX_N];

int score(int x, int y) {
	if (visit[x][y] == 1) return dp[x][y];
	visit[x][y] = 1;
	if (x == N - 1) return dp[x][y];
	return dp[x][y] += max(score(x + 1, y), score(x + 1, y + 1));
}

void solve() {
	for (int i = 0; i < N; i ++) {
		for (int j = 0; j <= i; j ++) {
			scanf("%d", &dp[i][j]);			
		}
	}
	printf("%d\n", score(0, 0));
}

int main() {
	while (~scanf("%d", &N)) {
		solve();
	}
	return 0;
}