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Description

Optimal Symmetric Paths

You have a grid of n rows and n columns. Each of the unit squares contains a non-zero digit. You walk from the top-left square to the bottom-right square. Each step, you can move left, right, up or down to the adjacent square (you cannot move diagonally), but you cannot visit a square more than once. There is another interesting rule: your path must be symmetric about the line connecting the bottom-left square and top-right square. Below is a symmetric path in a 6 x 6 grid.

Your task is to find out, among all valid paths, how many of them have the minimal sum of digits?

Input 

2n100

Output 

Sample Input 

2
1 1
1 1
3
1 1 1
1 1 1
2 1 1
0

Sample Output 

2
3

题意：求从左上角到右下角的最短路径数，且要求沿斜线对称
思路：既然要求对称，所以我们将对称的权值叠加，那么就是求到对角线的最短路径了，通过递推解决
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 111;
const int mod = 1000000009;

int n, num[maxn][maxn];
int dp[maxn][maxn];

void solve() {
	memset(dp, 0, sizeof(dp));
	for (int i = 1; i <= n; i++)
		dp[i][n-i+1] = 1;
	for (int k = n; k >= 2; k--)
		for (int i = k-1; i >= 1; i--) {
			int j = k - i;
			if (num[i+1][j] < num[i][j+1]) {
				num[i][j] += num[i+1][j];
				dp[i][j] = dp[i+1][j];
			}
			if (num[i+1][j] > num[i][j+1]) {
				num[i][j] += num[i][j+1];
				dp[i][j] = dp[i][j+1];
			}
			if (num[i+1][j] == num[i][j+1]) {
				num[i][j] += num[i+1][j];
				dp[i][j] = (dp[i+1][j] + dp[i][j+1]) % mod;
			}
		}
}

int main() {
	while (scanf("%d", &n) != EOF && n) {
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%d", &num[i][j]);
		for (int i = 1; i < n; i++)
			for (int j = 1; i+j <= n; j++)
				num[i][j] += num[n-j+1][n-i+1];
		solve();
		printf("%d\n", dp[1][1]%mod);
	}
	return 0;
}

UVA - 12295 Optimal Symmetric Paths （递推）,布布扣,bubuko.com

UVA - 12295 Optimal Symmetric Paths （递推）

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原文地址：http://blog.csdn.net/u011345136/article/details/38611871