ZOJ 2856 Happy Life
Do you know Utopia? It's a perfect world in which everyone leads a happy life.
A fairy wants to make a naive Utopia City. She studies the factors that have an impact on people's happiness and thinks that a person is happy if and only if all other persons' total "influence factor" on him (or her) is nonnegative. Each person has an influence factor on another person, which may be a positive integer or negative integer or 0. The influence factor is always symmetric, that is, if person A has an influence factor f on person B, it means that person B also has an "influence factor" f on A. So we can say the influence factor between person A and B is f without confusion. A person's influence factor on himself (or herself) is always 0. So let f(i, j) be the influence factor between person i and person j and person i's happiness(i) is well defined as follows:

Obviously there may be some persons who are not happy. Though the fairy cannot change any influence factor, she can give every person a property p(i) which is always +1 or -1. Under the fairy's magical definition, a person i's happiness(i)' is redefined as follow:

Person i feels happy if the value of happiness(i)' is nonnegative. But the fairy wonders whether she can give everyone a property to make all of them happy so that she can build her ideal naive Utopia successfully.
Since you're an ace programmer, the fairy asks you to help her to fulfill her dream. Can you help her?
Input
The input contains multiple test cases!
Each test case starts with an integer N (2 <= N <= 200), the number of persons in the city. After that there're N lines of integers and each line consists of N integers. The j-th integer of the i-th line of the matrix indicates the influence factor f(i,j) (-1000 < f(i,j) < 1000).
Proceed to the End Of File (EOF).
Output
For each test case, if the fairy fails, output a single line with "No" (without the quotations), otherwise output "Yes" (without the quotations) in the first line, followed by N lines, each line contains exactly a "+" (without the quotations) or a "-"(without the quotations) to indicate that the fairy should give the i-th person property +1 or -1 to fulfill her dream.
Sample Input
3
0 1 3
1 0 -1
3 -1 0
2
0 -10
-10 0
Sample Output
Yes
-
+
-
Yes
+
-
暴力,为什么不会出现No的情况呢?
#include <stdio.h>
#define MAXN 220 int n;
int g[MAXN][MAXN];
int ans[MAXN]; int main(){
int i,j;
while( scanf("%d",&n)!=EOF ){
for(i=1; i<=n; i++){
for(j=1; j<=n; j++){
scanf("%d" ,&g[i][j]);
}
ans[i]=1;
}
int index=1;
while(1){
if(index==n+1){
break;
}
int sum=0;
for(i=1; i<=n; i++){
sum+=ans[i]*g[index][i];
}
if(sum*ans[index]<0){
ans[index]*=-1;
index=1;
}
else index++;
}
puts("Yes");
for(int i=1; i<=n; i++){
if(ans[i]<0)
puts("-");
else
puts("+");
}
}
return 0;
}
ZOJ 2856 Happy Life的更多相关文章
- ZOJ 2856 Happy Life 暴力求解
因为是Special Judge 的题目,只要输出正确答案即可,不唯一 暴力力求解, 只要每次改变 happiness 值为负的人的符号即可. 如果计算出当前人的 happiness 值为负,那么将其 ...
- ZOJ People Counting
第十三届浙江省大学生程序设计竞赛 I 题, 一道模拟题. ZOJ 3944http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=394 ...
- ZOJ 3686 A Simple Tree Problem
A Simple Tree Problem Time Limit: 3 Seconds Memory Limit: 65536 KB Given a rooted tree, each no ...
- ZOJ Problem Set - 1394 Polar Explorer
这道题目还是简单的,但是自己WA了好几次,总结下: 1.对输入的总结,加上上次ZOJ Problem Set - 1334 Basically Speaking ac代码及总结这道题目的总结 题目要求 ...
- ZOJ Problem Set - 1392 The Hardest Problem Ever
放了一个长长的暑假,可能是这辈子最后一个这么长的暑假了吧,呵呵...今天来实验室了,先找了zoj上面简单的题目练练手直接贴代码了,不解释,就是一道简单的密文转换问题: #include <std ...
- ZOJ Problem Set - 1049 I Think I Need a Houseboat
这道题目说白了是一道平面几何的数学问题,重在理解题目的意思: 题目说,弗雷德想买地盖房养老,但是土地每年会被密西西比河淹掉一部分,而且经调查是以半圆形的方式淹没的,每年淹没50平方英里,以初始水岸线为 ...
- ZOJ Problem Set - 1006 Do the Untwist
今天在ZOJ上做了道很简单的题目是关于加密解密问题的,此题的关键点就在于求余的逆运算: 比如假设都是正整数 A=(B-C)%D 则 B - C = D*n + A 其中 A < D 移项 B = ...
- ZOJ Problem Set - 1001 A + B Problem
ZOJ ACM题集,编译环境VC6.0 #include <stdio.h> int main() { int a,b; while(scanf("%d%d",& ...
- zoj 1788 Quad Trees
zoj 1788 先输入初始化MAP ,然后要根据MAP 建立一个四分树,自下而上建立,先建立完整的一棵树,然后根据四个相邻的格 值相同则进行合并,(这又是递归的伟大),逐次向上递归 四分树建立完后, ...
随机推荐
- POJ 3581 Sequence(后缀数组)
Description Given a sequence, {A1, A2, ..., An} which is guaranteed A1 > A2, ..., An, you are to ...
- 5、Semantic-UI之基础按钮样式
5.1 基础按钮样式 在Semantic-UI中定义了很多的按钮样式,可以通过class="ui button"来指定,也可以在class中指定颜色. 示例:定义基础按钮样式 ...
- Android-动态添加控件到ScrollView
在实际开发过程中,会需要动态添加控件到ScrollView,就需要在Java代码中,找到ScrollView的孩子(ViewGroup),进行添加即可. Layout: <?xml versio ...
- [LeetCode 题解]: Maximum Subarray
前言 [LeetCode 题解]系列传送门: http://www.cnblogs.com/double-win/category/573499.html 1.题目描述 Find the c ...
- Ubuntu 网关服务器配置
1.设置Linux内核支持ip数据包的转发 echo "1" > /proc/sys/net/ipv4/ip_forward or vi /etc/sysctl.conf ...
- CSS基础知识:常见选择器示例
CSS(Cascading Style Sheet),中文译为层叠样式表,可以让设计者方便灵活地控制Web页面的外观表现.CSS是1996年由W3C审核通过并且推荐使用的.CSS的引入,就是为了使HT ...
- Verilog MIPS32 CPU(三)-- ALU
Verilog MIPS32 CPU(一)-- PC寄存器 Verilog MIPS32 CPU(二)-- Regfiles Verilog MIPS32 CPU(三)-- ALU Verilog M ...
- window.open之postMessage传参数
这次要实现一个window.open打开子视窗的同时传参数到子视窗,关闭的时候返回参数. 当然简单的做法非常简单,直接在window.open的URL之后接参数即可,但是毕竟get method的参数 ...
- IO--RAID
RAID IO计算 Raid 0 –每个磁盘的I/O计算= (读+写) /磁盘个数 Raid 1 --每个磁盘的I/O计算= [读+(2*写)]/2 Raid 5 --每个磁盘的I/O计算= [读+( ...
- 对路径“c:\windows\system32\inetsrv\syslog”的访问被拒绝。
win7 64 系统,在调试wcf的时候,出了这个错误,当时感觉iis的权限不够,iis搞了好长时间没解决.最后改了用到的应用程序池中的标识.标识改成 localSytem,之后问题解决. IIS-- ...