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
題目大意：
給出一個 n 層的數(shù)字三角形掏愁，要從最頂端走到最低端歇由，問哪條路徑上沿途的數(shù)字之和最大？輸出這個最大值
解題思路：
動態(tài)規(guī)劃法
用數(shù)組 a[][] 存儲數(shù)字三角形果港，定義數(shù)組 dp[][]沦泌，dp[i][j] 表示從 a[0][0] 走到 a[i][j] 能得到的最大值 顯然有 dp[0][0] = a[0][0]
我們要求的是max{dp[n-1][j]|0≤j≤n-1}
j=0 dp[i][0] = dp[i - 1][0] + a[i][0];
0＜j＜i dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j]) + a[i][j];
j=i dp[i][i] = dp[i - 1][i - 1] + a[i][i];
#include <stdio.h>
#include <iostream>
#define N 350
using namespace std;
int a[N][N], dp[N][N];
int max(int a, int b) {
return a>b ? a : b;
}
void init() {
memset(a, 0, sizeof(a));
memset(dp, 0, sizeof(dp));
}
int main()
{
int n;
int i, j;
int k = 1;
scanf("%d", &n);
init();
for (int i = 0; i < n; ++i)
{
for (j = 0;j<k;j++)
scanf("%d", &a[i][j]);
k++;
}
dp[0][0] = a[0][0];
for (i = 1; i < n; i++)
{
dp[i][0] = dp[i - 1][0] + a[i][0];
for (j = 1; j < i; j++)
dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j]) + a[i][j];
dp[i][i] = dp[i - 1][i - 1] + a[i][i];
}
int ans = 0;
for (int j = 0; j < n;j++)
ans = max(ans, dp[n - 1][j]);
printf("%d\n", ans);
return 0;
}
