UVA - 12295 最短路（迪杰斯特拉）——求按对称路线最短路条数

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cctype>
  4 #include<queue>
  5 #include<vector>
  6 #include <algorithm>
  7 using namespace std;
  8 const int maxn=105;
  9 const int INF=1e9+10;
 10 const int mod=1e9+9;
 11 typedef long long LL;
 12 int n,dp[maxn][maxn],w[maxn][maxn],counts[maxn][maxn];
 13 //dp[x][y]求的是从（0，0）点到（x，y）点的最短路径值（也就是最短路）
 14 //counts[x][y]求的是从（0，0）点到（x，y）点的最短路径数量
 15 int p[4][2]=
 16 {
 17     {1,0},
 18     {0,1},
 19     {-1,0},
 20     {0,-1}
 21 };
 22 struct shudui
 23 {
 24     int x,y,lx,ly,dis;
 25     shudui() {}
 26     shudui(int x,int y,int lx,int ly,int dis)
 27     {
 28         this->x=x;
 29         this->y=y;
 30         this->lx=lx;
 31         this->ly=ly;
 32         this->dis=dis;
 33     }
 34     bool operator < (const shudui a)const
 35     {
 36         return a.dis<dis;
 37     }
 38 } str1;
 39 priority_queue<shudui>r;
 40 void JK()
 41 {
 42     for(int i=0; i<maxn; ++i)
 43         fill(dp[i],dp[i]+maxn,mod);
 44     counts[0][0]=1;
 45     r.push(shudui(0,0,0,0,w[0][0]));
 46     while(!r.empty())
 47     {
 48         str1=r.top();
 49         r.pop();
 50         int x=str1.x;
 51         int y=str1.y;
 52         int lx=str1.lx;
 53         int ly=str1.ly;
 54         int dis=str1.dis;
 55         if(dp[x][y]>dis)
 56         {
 57             dp[x][y]=dis;
 58             counts[x][y]=counts[lx][ly];
 59         }
 60         else if(dp[x][y]==dis)
 61         {
 62             counts[x][y]=(counts[x][y]+counts[lx][ly])%mod;
 63             continue;
 64         }
 65         else continue;
 66 
 67         if(x+y>=n-1) continue;
 68         for(int i=0; i<4; ++i)
 69         {
 70             int xx=x+p[i][0];
 71             int yy=y+p[i][1];
 72             if(xx<n && yy<n && xx>=0 && yy>=0)
 73             {
 74                 r.push(shudui(xx,yy,x,y,dis+w[xx][yy]));
 75             }
 76         }
 77     }
 78 }
 79 int main()
 80 {
 81     while(~scanf("%d",&n) && n)
 82     {
 83         for(int i=0; i<n; ++i)
 84         {
 85             for(int j=0; j<n; ++j)
 86             {
 87                 scanf("%d",&w[i][j]);
 88             }
 89         }
 90         for(int i = 0; i < n; i++)
 91         {
 92             for(int j = 0; j < n-i-1; j++)
 93             {
 94                 w[i][j] += w[n-j-1][n-i-1];
 95             }
 96         }
 97         JK();
 98         int minn=mod;
 99         for(int i=0; i<n; ++i)
100         {
101             minn=min(minn,dp[i][n-i-1]);
102         }
103         int ans=0;
104         for(int i=0; i<n; ++i)
105         {
106             if(dp[i][n-i-1]==minn)
107             {
108                 ans=(ans+counts[i][n-i-1])%mod;
109             }
110         }
111         printf("%d\n",ans);
112     }
113     return 0;
114 }