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
- java中super的几种用法，与this的区别
1. 子类的构造函数如果要引用super的话,必须把super放在函数的首位. class Base { Base() { System.out.println("Base"); ...
- Linux基础命令---ab测试apache性能
ab ab指令是apache的性能测试工具,它可以测试当前apache服务器的运行性能,显示每秒中可以处理多少个http请求. 此命令的适用范围:RedHat.RHEL.Ubuntu.CentOS.F ...
- 类型类 && .class 与 .getClass() 的区别
一. 什么是类型类 Java 中的每一个类(.java 文件)被编译成 .class 文件的时候,Java虚拟机(JVM)会为这个类生成一个类对象(我们姑且认为就是 .class 文件),这个对象包含 ...
- 【Linux】【Shell】【Basic】数组
1. 数组: 变量:存储单个元素的内存空间: 数组:存储多个元素的连续的内存空间: 数组名:整个数组只有一个名字: 数组 ...
- Docker常用image
MySQL Start a mysql server instance Starting a MySQL instance is simple: docker run -itd --name mysq ...
- linux 操作只读变量
由于该操作需要用到 gdb,所以需要先 安装好 gdb 1. 查询是否有gdb: 2. 如果没有,需要先执行 yum install gdb 命令进行安装 3. 定义 只读变量 abc 并打印值: a ...
- AOP中环绕通知的写法
package com.hope.utils;import org.aspectj.lang.ProceedingJoinPoint;/** * @author newcityman * @date ...
- 【C/C++】最长不下降子序列/动态规划
#include <iostream> #include <vector> using namespace std; int main() { //输入 int tmp; ve ...
- 微服务下前后端分离的统一认证授权服务，基于Spring Security OAuth2 + Spring Cloud Gateway实现单点登录
1. 整体架构 在这种结构中,网关就是一个资源服务器,它负责统一授权(鉴权).路由转发.保护下游微服务. 后端微服务应用完全不用考虑权限问题,也不需要引入spring security依赖,就正常的 ...
- NGNIX 开启socket分发的使用配置
worker_processes 1; #error_log logs/error.log; #error_log logs/error.log notice; #error_log logs/err ...