poj3176(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 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

最基础的数塔dp了，维数可以降到一维来优化空间，不过这样要注意循环顺序，以免覆盖之前结果

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double EPS = 1e-8;
const int MAXN = 360;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};

int main()
{
    int n;
    int s[MAXN][MAXN], dp[MAXN];
    cin >> n;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j <= i; ++j)
            scanf("%d", &s[i][j]);
    for (int i = 0; i < n; ++i)
        dp[i] = s[n-1][i];
    for (int i = n-2; i >= 0; --i)
        for (int j = 0; j <= i; ++j)
            dp[j] = max(dp[j], dp[j+1]) + s[i][j];
    cout << dp[0] << endl;
    return 0;
}