uva 12295 - Optimal Symmetric Paths(最短路 + 递推)

  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.


\epsfbox{p12295.eps}
Your task is to find out, among all valid paths, how many of them have the minimal sum of digits?


Input 


There will be at most 25 test cases. Each test case begins with an integer n ( 2$ \le$n$ \le$100). Each of the next n lines contains n non-zero digits (i.e. one of 1, 2, 3, ..., 9). These n2 integers are the digits in the grid. The input is terminated by a test case with n = 0, you should not process it.
Output 


For each test case, print the number of optimal symmetric paths, modulo 1,000,000,009.
Sample Input 


2
1 1
1 1
3
1 1 1
1 1 1
2 1 1
0
Sample Output 


2
3




The Seventh Hunan Collegiate Programming Contest 

Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan


找从左上角到右下角的,按y=x对称的路径,最短路径有多少条。
因为是对称的,所以只用找到y=x上的点的路径,可以先按对称性做一个预处理,把对称点的digit加起来。最后做最短路计数。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 100 + 5;
const int mod = 1000000000 + 9;

int m[maxn][maxn];
int dp[maxn][maxn];
int countn[maxn][maxn];
int n;
const int cx[] = {-1, 0, 1, 0};
const int cy[] = {0, 1, 0, -1};

bool valid(int x, int y){
    return x < n && x >= 0 && y < n && y >= 0;
}

struct Node{
    int x, y, dis;
    int lx, ly;

    Node(int x, int y, int dis, int lx, int ly){
        this->x = x;
        this->y = y;
        this->dis = dis;
        this->lx = lx;
        this->ly = ly;
    }

    bool operator < (const Node &a) const{
        return a.dis < dis;
    }
};

void solve(){
    priority_queue<Node> pq;
    for(int i = 0;i < maxn;i++)
        fill(dp[i], dp[i]+maxn, mod);
    countn[0][0] = 1;
    pq.push(Node(0, 0, m[0][0], 0, 0));// !!!
    while(!pq.empty()){
        Node tem = pq.top();pq.pop();
        int x = tem.x;
        int y = tem.y;
        int dis = tem.dis;
        int lx = tem.lx;
        int ly = tem.ly;

        if(dis < dp[x][y]){
            dp[x][y] = dis;
            countn[x][y] = countn[lx][ly];
        }
        else if(dis == dp[x][y]){
            countn[x][y] = (countn[x][y] + countn[lx][ly]) % mod;
            continue;
        }
        else if(dis > dp[x][y]) continue;

        if(x + y == n-1) continue;
        for(int i = 0;i < 4;i++){
            int tx = x + cx[i];
            int ty = y + cy[i];
            if(valid(tx, ty)){
                pq.push(Node(tx, ty, dis+m[tx][ty], x, y));
            }
        }
    }
}

int main(){
    while(scanf("%d", &n) != EOF){
        if(n == 0) break;
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                scanf("%d", &m[i][j]);
            }
        }
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n-i-1;j++){
                m[i][j] += m[n-j-1][n-i-1];
            }
        }

        solve();
        int Min = mod;
        for(int i = 0;i < n;i++){
            Min = min(Min, dp[i][n-i-1]);
        }
        int ans = 0;
        for(int i = 0;i < n;i++){
            if(dp[i][n-i-1] == Min)
                ans = (ans + countn[i][n-i-1]) % mod;
        }
        printf("%d\n", ans);
    }
    return 0;
}


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