POJ 3176 DP（简单数塔）

Cow Bowling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 15863 Accepted: 10567

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   02   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

题意：
数塔从上到下，每下一层只能选和上一个位置左右相邻的位置，问怎么走时经过的路径权重相加最大。
解法：
比较经典的dp了，之前在hdu做了一次。这次做做要是求速度，应该是几分钟就敲完AC了。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define F(i,a,b) for(int i = a;i<=b;i++)#define FI(i,a,b) for(int i = a;i>=b;i--)int dp[1000];int G[1000][1000];using namespace std;int main(){//    freopen("data.in","r",stdin);    int n;    while(scanf("%d",&n)!=EOF){        F(i,1,n)            F(j,1,i)                scanf("%d",&G[i][j]);        F(i,1,n)            dp[i] = G[n][i];        FI(i,n-1,1)            F(j,1,i)                dp[j] = G[i][j] + max(dp[j],dp[j+1]);        printf("%d\n",dp[1]);    }    return 0;}