poj_1163

题目：

Description

7
3   8
8   1   0
2   7   4   4
4   5   2   6   5

(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.
Output

Your program is to write to standard output. The highest sum is written as an integer.
Sample Input

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output

30

分析：一个dp问题，首先写出状态定义，这里可以用d(x,y)表示二维数组中row = x, col = y 的结点的到最底层（row最大）时的和的最大值，那么问题就可以转化为求d(1,1)(我是把数据从1开始放置的)；可以写出状态转移方程：d(x, y) = v(x, y) + max(d(x+1, y), d(x+1, y+1));

代码如下：

#include<stdio.h>

int v[60][60];
int d[60][60];
int main() {
    int n, i, j, max;
    scanf("%d", &n);
    for(i = 1; i <= n; ++i) {
        for(j = 1; j <= i; ++j) {
            scanf("%d", &v[i][j]);
        }
    }

    for(i = n; i >= 1; --i){
        for(j = 1; j <= i; ++j) {
            if(i == n) {
                d[i][j] = v[i][j];
            } else {
                max = (d[i+1][j] > d[i+1][j+1] ? d[i+1][j] : d[i+1][j+1]);
                d[i][j] = max + v[i][j];
            }
        }
    }
    printf("%d", d[1][1]);
}