poj 3176 -- Cow Bowling

Cow Bowling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13931		Accepted: 9230

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

---

题目链接：Cow Bowling

思路:简单dp，白皮书入门dp。

/*======================================================================
 *           Author :   kevin
 *         Filename :   CowBowing.cpp
 *       Creat time :   2014-09-18 10:59
 *      Description :
========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 400
using namespace std;
int s[M][M],dp[M][M];
int main(int argc,char *argv[])
{
    int n;
    while(scanf("%d",&n)!=EOF){
        clr(dp,0);
        clr(s,0);
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= i; j++){
                scanf("%d",&s[i][j]);
            }
        }
        for(int i = 1; i <= n; i++){
            dp[n][i] = s[n][i];
        }
        for(int i = n-1; i >= 1; i--){
            for(int j = 1; j <= i; j++){
                dp[i][j] = max(s[i][j] + dp[i+1][j],s[i][j] + dp[i+1][j+1]);
            }
        }
        printf("%d\n",dp[1][1]);
    }
    return 0;
}