Optimal Symmetric Paths(UVA12295)

最新推荐文章于 2020-06-08 17:16:56 发布

转载最新推荐文章于 2020-06-08 17:16:56 发布 · 68 阅读

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原文链接：http://www.cnblogs.com/zzuli2sjy/p/5656023.html

本文介绍了一种算法问题，即在一个由非零数字组成的网格中寻找所有关于特定线对称的有效路径中，数字之和最小的路径数量。通过将网格划分为对称的两部分并利用Dijkstra算法来寻找最短路径，最终统计这些最短路径的数量。

Description

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

Sample Output

2
3
思路：要求是要关于那条线对称的，所一把上半角和下半角叠加起来，然后求到那条线的最短路即可，用迪杰斯特拉求。建图也比较简单,就是每个点向四个方向的点连边
。在求最短路的时候开一个数组记录当前走到该点的最短路有多少条就行，最后求到斜边点上等于最短路的种数和即可。
复杂度n*n

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<math.h>
  6 #include<queue>
  7 #include<vector>
  8 using namespace std;
  9 int ma[200][200];
 10 typedef struct pp
 11 {
 12         int x;
 13         int y;
 14         int cost;
 15         int id;
 16         bool flag;
 17 } ss;
 18 const int mod=1e9+9;
 19 typedef long long LL;
 20 LL sum[10005];
 21 LL  d[10005];
 22 bool flag[10005];
 23 ss node[10005];
 24 vector<ss>vec[10005];
 25 int dd[200][200];
 26 void dj(int n,int id);
 27 int main(void)
 28 {
 29         int i,j,k;
 30         while(scanf("%d",&k),k!=0)
 31         {
 32                 memset(dd,-1,sizeof(dd));
 33                 memset(flag,0,sizeof(flag));
 34                 memset(sum,0,sizeof(sum));
 35                 for(i=0;i<10005;i++)
 36                     vec[i].clear();
 37                 for(i=0; i<k; i++)
 38                 {
 39                         for(j=0; j<k; j++)
 40                         {
 41                                 scanf("%d",&ma[i][j]);
 42                         }
 43                 }
 44                 for(i=0; i<k; i++)
 45                 {
 46                         for(j=0; j<(k-i); j++)
 47                         {
 48                                 if(i+j!=k-1)
 49                                 {
 50                                         ma[i][j]+=ma[k-j-1][k-i-1];
 51                                 }
 52                         }
 53                 }
 54                 int id=0;
 55                 for(i=0; i<k; i++)
 56                 {
 57                         for(j=0; j<(k-i); j++)
 58                         {
 59                                 if(i+j==k-1)
 60                                 {
 61                                         node[id].flag=true;
 62                                         node[id].x=i;
 63                                         node[id].y=j;
 64                                         node[id].id=id;
 65                                 }
 66                                 else
 67                                 {
 68                                         node[id].flag=false ;
 69                                         node[id].x=i;
 70                                         node[id].y=j;
 71                                         node[id].id=id;
 72                                 }
 73                                 dd[i][j]=id;
 74                                 if(i-1>=0)
 75                                 {
 76                                         ss cc;
 77                                         cc.x=i-1;
 78                                         cc.y=j;
 79                                         cc.id=dd[i-1][j];
 80                                         cc.cost=ma[i-1][j];
 81                                         vec[id].push_back(cc);
 82                                         cc.x=i;
 83                                         cc.y=j;
 84                                         cc.id=dd[i][j];
 85                                         cc.cost=ma[i][j];
 86                                         vec[dd[i-1][j]].push_back(cc);
 87                                 }
 88                                 if(j-1>=0)
 89                                 {
 90                                         ss cc;
 91                                         cc.x=i;
 92                                         cc.y=j-1;
 93                                         cc.id=dd[i][j-1];
 94                                         cc.cost=ma[i][j-1];
 95                                         vec[id].push_back(cc);
 96                                         cc.x=i;
 97                                         cc.y=j;
 98                                         cc.id=dd[i][j];
 99                                         cc.cost=ma[i][j];
100                                         vec[dd[i][j-1]].push_back(cc);
101                                 }
102                                 if(i+1<k&&dd[i+1][j]!=-1)
103                                 {
104                                         ss cc;
105                                         cc.x=i+1;
106                                         cc.y=j;
107                                         cc.id=dd[i+1][j];
108                                         cc.cost=ma[i+1][j];
109                                         vec[id].push_back(cc);
110                                         cc.x=i;
111                                         cc.y=j;
112                                         cc.id=dd[i][j];
113                                         cc.cost=ma[i][j];
114                                         vec[dd[i+1][j]].push_back(cc);
115                                 }
116                                 if(j+1<k&&dd[i][j+1]!=-1)
117                                 {
118                                         ss cc;
119                                         cc.x=i;
120                                         cc.y=j+1;
121                                         cc.id=dd[i][j+1];
122                                         cc.cost=ma[i][j+1];
123                                         vec[id].push_back(cc);
124                                         cc.x=i;
125                                         cc.y=j;
126                                         cc.id=dd[i][j];
127                                         cc.cost=ma[i][j];
128                                         vec[dd[i][j+1]].push_back(cc);
129                                 }
130                                 id++;
131                         }
132                 }
133                 dj(0,id);
134                 LL maxx=1e18;
135                 for(i=0; i<id; i++)
136                 {
137                         if(node[i].flag)
138                         {
139                                 if(maxx>d[i])
140                                 {
141                                         maxx=d[i];
142                                 }
143                         }
144                 }
145                 LL akk=0;
146                 for(i=0; i<id; i++)
147                 {
148                         if(maxx==d[i]&&node[i].flag)
149                         {
150                                 akk=akk+sum[i];
151                                 akk%=mod;
152                         }
153                 }
154                 printf("%lld\n",akk);
155         }
156         return 0;
157 }
158 void dj(int n,int id)
159 {
160         int i,j,k;
161         fill(d,d+10005,1e9);
162         d[n]=ma[0][0];
163         memset(flag,0,sizeof(flag));
164         while(true)
165         {
166                 int l=-1;
167                 for(i=0; i<id; i++)
168                 {
169                         if((l==-1||d[i]<d[l])&&flag[i]==false)
170                         {
171                                 l=i;
172                         }
173                 }
174                 if(l==-1)
175                 {
176                         return ;
177                 }
178                 flag[l]=true;
179                 ss   ask=node[l];
180                 int x=ask.x;
181                 int y=ask.y;
182                 int ac=ask.id;
183                 if(l==0)
184                 {
185                         sum[l]=1;
186                 }
187                 else
188                 {
189 
190                         for(i=0; i<vec[ac].size(); i++)
191                         {
192                                 ss pp=vec[ac][i];
193                                 if(d[pp.id]+(LL)ma[x][y]==d[l])
194                                         sum[l]=sum[pp.id]+sum[l];
195                                 sum[l]%=mod;
196                         }
197                 }
198                 for(i=0; i<vec[ac].size(); i++)
199                 {
200                         ss pp=vec[ac][i];
201                         if(d[pp.id]>d[l]+pp.cost)
202                                 d[pp.id]=d[l]+pp.cost;
203                 }
204         }
205 }

转载于:https://www.cnblogs.com/zzuli2sjy/p/5656023.html