LA 3713
The Bandulu Space Agency (BSA) has plans for the following three space missions:
- Mission A: Landing on Ganymede, the largest moon of Jupiter.
- Mission B: Landing on Callisto, the second largest moon of Jupiter.
- Mission C: Landing on Titan, the largest moon of Saturn.
Your task is to assign a crew for each mission. BSA has trained a number of excellent astronauts; everyone of them can be assigned to any mission. However, if two astronauts hate each other, then it is not wise to put them on the same mission. Furthermore, Mission A is clearly more prestigious than Mission B; who would like to go to the second largest moon if there is also a mission to the largest one? Therefore, the assignments have to be done in such a way that only young, inexperienced astronauts go to Mission B, and only senior astronauts are assigned to Mission A. An astronaut is considered young if their age is less than the average age of the astronauts and an astronaut is senior if their age is at least the averageage. Every astronaut can be assigned to Mission C, regardless of their age (but you must not assign two astronauts to the same mission if they hate each other).
Input
The input contains several blocks of test cases. Each case begins with a line containing two integers 1n
100000 <tex2html_verbatim_mark>and 1
m
100000 <tex2html_verbatim_mark>. The number n <tex2html_verbatim_mark>is the number of astronauts. The next n <tex2html_verbatim_mark>lines specify the age of the n<tex2html_verbatim_mark>astronauts; each line contains a single integer number between 0 and 200. The next m <tex2html_verbatim_mark>lines contains two integers each, separated by a space. A line containing i <tex2html_verbatim_mark>and j <tex2html_verbatim_mark>(1
i, j
n) <tex2html_verbatim_mark>means that the i <tex2html_verbatim_mark>-th astronaut and the j <tex2html_verbatim_mark>-th astronaut hate each other.
The input is terminated by a block with n = m = 0 <tex2html_verbatim_mark>.
Output
For each test case, you have to output n lines, each containing a single letter. This letter is either `A', `B', or `C'. The i <tex2html_verbatim_mark>-th line describes which mission the i <tex2html_verbatim_mark>-th astronaut is assigned to. Astronauts that hate each other should not be assigned to the same mission, only young astronauts should be assigned to Mission B and only senior astronauts should be assigned to Mission A. If there is no such assignment, then output the single line `No solution.' (without quotes).
Sample Input
16 20
21
22
23
24
25
26
27
28
101
102
103
104
105
106
107
108
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
1 10
2 9
3 12
4 11
5 14
6 13
7 16
8 15
1 12
1 13
3 16
6 15
0 0
Sample Output
B
C
C
B
C
B
C
B
A
C
C
A
C
A
C
A 拆成两个分组走2-sat.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <vector> using namespace std; const int MAX_N = 1e5 + ;
int N,M;
int low[MAX_N * ],pre[MAX_N * ],cmp[MAX_N * ];
int age[MAX_N];
vector <int> G[ * MAX_N];
stack <int> S;
int dfs_clock,scc_cnt; void dfs(int u) {
pre[u] = low[u] = ++dfs_clock;
S.push(u);
for(int i = ; i < G[u].size(); ++i) {
int v = G[u][i];
if(!pre[v]) {
dfs(v);
low[u] = min(low[u],low[v]);
} else if(!cmp[v]) {
low[u] = min(low[u],pre[v]);
}
} if(pre[u] == low[u]) {
++scc_cnt;
for(;;) {
int x = S.top(); S.pop();
cmp[x] = scc_cnt;
if(x == u) break;
}
}
} bool scc() {
dfs_clock = scc_cnt = ;
memset(cmp,,sizeof(cmp));
memset(pre,,sizeof(pre)); for(int i = ; i <= * N; ++i) {
if(!pre[i]) dfs(i);
} for(int i = ; i <= N; ++i) {
if(cmp[i] == cmp[N + i]) return false;
} return true;
}
int main()
{
freopen("sw.in","r",stdin);
while(~scanf("%d%d",&N,&M) && (N || M)) {
int sum = ;
for(int i = ; i <= N; ++i) {
scanf("%d",&age[i]);
sum += age[i];
}
for(int i = ; i <= * N; ++i) G[i].clear(); for(int i = ; i <= M; ++i) {
int u,v;
scanf("%d%d",&u,&v);
if(age[u] * N >= sum && age[v] * N < sum
|| age[u] * N < sum && age[v] * N >= sum) {
G[v + N].push_back(u);
G[u + N].push_back(v);
} else { G[u].push_back(v + N);
G[v].push_back(u + N);
G[v + N].push_back(u);
G[u + N].push_back(v);
} } if(!scc()) {
printf("No solution.\n");
} else {
for(int i = ; i <= N; ++i) {
if(age[i] * N >= sum) {
printf("%c\n",cmp[i] < cmp[i + N] ? 'A' : 'C');
} else {
printf("%c\n",cmp[i] < cmp[i + N] ? 'B' : 'C');
}
}
}
} return ;
}
LA 3713的更多相关文章
- LA 3713 宇航员分组
题目链接:http://vjudge.net/contest/142615#problem/B 题意:有A,B,C三个人物要分配个N个宇航员,每个宇航员恰好要分配一个任务,设平均年龄为X,只有年龄大于 ...
- LA 3713 Astronauts
给个题目链接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=sh ...
- 图论$\cdot$2-SAT问题
2-SAT问题是这样的:有$n$个布尔变量$x_i$,另有$m$个需要满足的条件,每个条件的形式都是“$x_i$为真/假或者$x_j$为真/假”.比如:"$x_1$为真或者$x_3$为假“. ...
- 2-SAT 问题与解法小结
2-SAT 问题与解法小结 这个算法十分的奇妙qwq... 将一类判定问题转换为图论问题,然后就很容易解决了. 本文有一些地方摘录了一下赵爽<2-SAT解法浅析> (侵删) 一些概念: \ ...
- [UOJ317]【NOI2017】游戏 题解
题意 小 L 计划进行 \(n\) 场游戏,每场游戏使用一张地图,小 L 会选择一辆车在该地图上完成游戏. 小 L 的赛车有三辆,分别用大写字母 A.B.C 表示.地图一共有四种,分别用小写字 ...
- leggere la nostra recensione del primo e del secondo
La terra di mezzo in trail running sembra essere distorto leggermente massima di recente, e gli aggi ...
- Le lié à la légèreté semblait être et donc plus simple
Il est toutefois vraiment à partir www.runmasterfr.com/free-40-flyknit-2015-hommes-c-1_58_59.html de ...
- Mac Pro 使用 ll、la、l等ls的别名命令
在 Linux 下习惯使用 ll.la.l 等ls别名的童鞋到 mac os 可就郁闷了~~ 其实只要在用户目录下建立一个脚本“.bash_profile”, vim .bash_profile 并输 ...
- Linux中的动态库和静态库(.a/.la/.so/.o)
Linux中的动态库和静态库(.a/.la/.so/.o) Linux中的动态库和静态库(.a/.la/.so/.o) C/C++程序编译的过程 .o文件(目标文件) 创建atoi.o 使用atoi. ...
随机推荐
- webpack 学习笔记 02 快速入门
webpack 的目标 将依赖项分块,按需加载. 减少web app的初始加载时间. 使每一个静态集合都能够作为组件使用. 有能力集成第三方库,作为组件使用. 高度可配置化. 适用于大型项目. INS ...
- mysql外键设置(待测试)
外键的定义语法:[CONSTRAINT symbol] FOREIGN KEY [id] (index_col_name, ...) REFERENCES tbl_name (index_col ...
- Drawable
今天简单的介绍一下有关以下5中的应用: Statelistdrawable Layerdrawable Shapeddrawable Clipdrawable Animationdrawable 1. ...
- hdu 4308 Saving Princess claire_
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4308 Saving Princess claire_ Description Princess cla ...
- ORA-01017 invalid username/password;logon denied" (密码丢失解决方案)
1.先确认是否输错 用户名和密码 2.如果的确是丢失密码的话: 查看sqlnet.ora 如果是 SQLNET.AUTHENTICATION_SERVICES= (NONE) , 需更改为SQLNET ...
- XAML特殊字符
此部分只限制在XAML中,代码中不受此类限制. 1.特殊字符转义 XAML 特殊字符转义 特殊字符 转义 小于号 < < 大于号 > > 取址符 & & 引号 ...
- libevent I/O示例
I/O示例使用一个windows平台上服务器/客户端的例子来演示.由于为了减少代码篇幅等各种由于本人懒而产生的原因,以下代码没有做错误处理以及有些小问题,但是我想应该不影响演示,大家多包涵. 服务器代 ...
- c++基础(二):成员he派生类
struct Date{ int day, month, year; }; struct Book{ string name; Date date; void init(string n, int y ...
- linux系统下sd卡的备份与恢复
linux系统下sd卡的备份与恢复 现在各种的开发板都是从sd卡上面启动的,因此大修改工作之前很有必要备份一下. 备份 在linux系统下用读卡器读取sd卡 用df -h命令看分区的路径 一般都是/d ...
- skill-判断浏览器
判断是ie浏览器还是火狐等标准浏览器 var ie=!+"\v1"; 因为ie浏览器不支持\v,也就是水平制表符,所以"\"符号会被忽略,前面的+号是把&quo ...