P OJ-----3176DP动态规划入门

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17407 Accepted: 11611

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

573 88 1 02 7 4 44 5 2 6 5

Sample Output

30

数字组成三角形，像杨辉三角一样，问从上到下只能走下方相邻的两个中的一个，所有数字之和最大是多少，另开一个数组储存每层的最大值，从下往上计算，最后最上方的那个就是最大值

#include<cstdio>#include<algorithm>using namespace std;int s[360][360], ss[360][360];int main(){int n;scanf("%d", &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++){ss[n-1][i] = s[n-1][i];}for(int i = n-2; i >= 0; i--){for(int j = 0; j <= i; j++){ss[i][j] = s[i][j] + max(ss[i+1][j], ss[i+1][j+1]);}}printf("%d\n", ss[0][0]);return 0;}

优化

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int map[360][360];int main(){int t, n;scanf("%d", &t);while(t--){memset(map, 0, sizeof(map));scanf("%d", &n);for(int i = 0; i < n; i++){for(int j = 0; j <= i; j++){scanf("%d", &map[i][j]);}}for(int i = n-2; i >= 0; i--){for(int j = 0; j <= i; j++){map[i][j] = max(map[i+1][j], map[i+1][j+1]) + map[i][j];}}printf("%d\n", map[0][0]);}return 0;}