题目链接：POJ 3176 Cow Bowling
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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 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 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. Source |
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;
}