Optimal Symmetric Paths(UVA12295)
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.
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
n100). 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
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 }
Optimal Symmetric Paths(UVA12295)的更多相关文章
- FAQ: Automatic Statistics Collection (文档 ID 1233203.1)
In this Document Purpose Questions and Answers What kind of statistics do the Automated tasks ...
- Contest2073 - 湖南多校对抗赛（2015.04.06）
Contest2073 - 湖南多校对抗赛(2015.04.06) Problem A: (More) Multiplication Time Limit: 1 Sec Memory Limit: ...
- Optimal Milking 分类： 图论 POJ 最短路 查找 2015-08-10 10:38 3人阅读 评论(0) 收藏
Optimal Milking Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 13968 Accepted: 5044 Case ...
- POJ2112 Optimal Milking （网络流）（Dinic）
Optimal Milking Time Limit: 2000MS Memory Limit: 30000K T ...
- POJ 2112 Optimal Milking （二分+最短路径+网络流）
POJ 2112 Optimal Milking (二分+最短路径+网络流) Optimal Milking Time Limit: 2000MS Memory Limit: 30000K To ...
- POJ 2112 Optimal Milking （Dinic + Floyd + 二分）
Optimal Milking Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 19456 Accepted: 6947 ...
- POJ2112 Optimal Milking
Optimal Milking Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 17811 Accepted: 6368 ...
- Optimal Milking POJ - 2112 （多重最优匹配+最小费用最大流+最大值最小化 + Floyd）
Optimal Milking Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 19347 Accepted: 690 ...
- POJ2112：Optimal Milking(Floyd+二分图多重匹配+二分)
Optimal Milking Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 20262 Accepted: 7230 ...
随机推荐
- leetcode刷题之数组NO.4
1.题目 给定一个数组 nums,编写一个函数将所有 0 移动到数组的末尾,同时保持非零元素的相对顺序. 示例: 输入: [0,1,0,3,12] 输出: [1,3,12,0,0] 说明: 必须在原数 ...
- c#页面查询、数据显示
page : <%@ Control Language="C#" AutoEventWireup="true" CodeFile="Queryx ...
- mysql死锁com.mysql.cj.jdbc.exception.MYSQLTransactionRollbackException Deadlock found when trying to get lock;try restarting transaction
1.生产环境出现以下报错 该错误发生在update操作中,该表并未建立索引,也就是只有InnoDB默认的主键索引,发生错误的程序是for循环中update. 什么情况下会出现Deadlock foun ...
- springmvc中拦截器的定义和配置
package com.hope.interceptor;import org.springframework.lang.Nullable;import org.springframework.web ...
- JSP中session、cookie和application的使用
一.session (单用户使用) 1.用处:注册成功后自动登录,登录后记住用户状态等 使用会话对象session实现,一次会话就是一次浏览器和服务器之间的通话,会话可以在多次请求中保存和使用数据. ...
- Mysql资料 查询条件
目录 一.计算 二.比较 三.逻辑运算符 四.位运算符 五.优先顺序 一.计算 二.比较 三.逻辑运算符 四.位运算符 五.优先顺序 实际上,很少有人能将这些优先级熟练记忆,很多情况下我们都是用&qu ...
- 《转》谈谈基于Kerberos的Windows Network Authentication
http://www.cnblogs.com/artech/archive/2007/07/05/807492.html 基本原理引入Key Distribution: KServer-Client从 ...
- Windwos堆管理体系以及溢出利用
<0day安全>学习笔记,主要讨论WIndows2000~WIndowsSP1平台的堆管理策略. 0X01 堆与栈的区别 栈空间是在程序设计时已经规定好怎么使用,使用多少内存空间.典型的栈 ...
- AJAX get和post请求
<!DOCTYPE html><html><head> <meta charset="UTF-8"> <title>&l ...
- webservice注意事项
1.private static final QName PORT_NAME = new QName("http://server.helloworld.cxf.demo/",&q ...