bzoj1681[Usaco2005 Mar]Checking an Alibi 不在场的证明
Description
A crime has been comitted: a load of grain has been taken from the barn by one of FJ's cows. FJ is trying to determine which of his C (1 <= C <= 100) cows is the culprit.
Fortunately, a passing satellite took an image of his farm M (1 <= M <= 70000) seconds before the crime took place, giving the location of all of the cows. He wants to know which cows had time to get to the barn to steal the grain. Farmer John's farm comprises
F (1 <= F <= 500) fields numbered 1..F and connected by P (1 <= P <= 1,000) bidirectional paths whose traversal time is in the range 1..70000 seconds (cows walk very slowly). Field 1 contains the barn. It takes no time to travel within a field (switch paths).
Given the layout of Farmer John's farm and the location of each cow when the satellite flew over, determine set of cows who could be guilty. NOTE: Do not declare a variable named exactly 'time'. This will reference the system call and never give you the results
you really want.
Input
* Line 1: Four space-separated integers: F, P, C, and M * Lines 2..P+1: Three space-separated integers describing a path: F1, F2, and T. The path connects F1 and F2 and requires T seconds to traverse.
* Lines P+2..P+C+1: One integer per line, the location of a cow. The first line gives the field number of cow 1, the second of cow 2, etc.
第1行输入四个整数F,只C,和M;
接下来P行每行三个整数描述一条路,起点终点和通过时间.
Output
* Line 1: A single integer N, the number of cows that could be guilty of the crime.
* Lines 2..N+1: A single cow number on each line that is one of the cows that could be guilty of the crime. The list must be in ascending order.
Sample Input
1 4 2
1 2 1
2 3 6
3 5 5
5 4 6
1 7 9
1
4
5
3
7
Sample Output
1
2
3
4
HINT
因为数据实在太弱,各种最短路都能过……
写了个floyd
#include<cstring>
#include<cstdio>
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline int min(int a,int b) {if(a<b)return a;else return b;}
int n,m,c,t,len;
int dist[1001][1001];
int ans[1001];
int main()
{
n=read();
m=read();
c=read();
t=read();
memset(dist,127/3,sizeof(dist));
for (int i=1;i<=n;i++)dist[i][i]=0;
for(int i=1;i<=m;i++)
{
int x=read(),y=read(),z=read();
if (z<=t)
{
dist[x][y]=min(dist[x][y],z);
dist[y][x]=dist[x][y];
}
}
for (int k=1;k<=n;k++)
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
if (i!=j)
if (dist[i][k]+dist[k][j]<dist[i][j])
dist[i][j]=dist[i][k]+dist[k][j];
for (int i=1;i<=c;i++)
{
int x=read();
if (dist[x][1]>t) continue;
ans[++len]=i;
}
for (int i=1;i<len;i++)
for (int j=i+1;j<=len;j++)
if(ans[i]>ans[j])
{
int t=ans[i];
ans[i]=ans[j];
ans[j]=t;
}
printf("%d\n",len);
for (int i=1;i<=len;i++)
printf("%d\n",ans[i]);
}
bzoj1681[Usaco2005 Mar]Checking an Alibi 不在场的证明的更多相关文章
- bzoj:1681 [Usaco2005 Mar]Checking an Alibi 不在场的证明
Description A crime has been comitted: a load of grain has been taken from the barn by one of FJ's c ...
- 【BZOJ】1681: [Usaco2005 Mar]Checking an Alibi 不在场的证明(spfa)
http://www.lydsy.com/JudgeOnline/problem.php?id=1681 太裸了.. #include <cstdio> #include <cstr ...
- BZOJ1680: [Usaco2005 Mar]Yogurt factory
1680: [Usaco2005 Mar]Yogurt factory Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 106 Solved: 74[Su ...
- BZOJ 1739: [Usaco2005 mar]Space Elevator 太空电梯
题目 1739: [Usaco2005 mar]Space Elevator 太空电梯 Time Limit: 5 Sec Memory Limit: 64 MB Description The c ...
- BZOJ 1738: [Usaco2005 mar]Ombrophobic Bovines 发抖的牛( floyd + 二分答案 + 最大流 )
一道水题WA了这么多次真是.... 统考终于完 ( 挂 ) 了...可以好好写题了... 先floyd跑出各个点的最短路 , 然后二分答案 m , 再建图. 每个 farm 拆成一个 cow 点和一个 ...
- 1740: [Usaco2005 mar]Yogurt factory 奶酪工厂
1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 119 Solved: ...
- 1682: [Usaco2005 Mar]Out of Hay 干草危机
1682: [Usaco2005 Mar]Out of Hay 干草危机 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 391 Solved: 258[ ...
- bzoj1682[Usaco2005 Mar]Out of Hay 干草危机*
bzoj1682[Usaco2005 Mar]Out of Hay 干草危机 题意: 给个图,每个节点都和1联通,奶牛要从1到每个节点(可以走回头路),希望经过的最长边最短. 题解: 求最小生成树即可 ...
- bzoj1680[Usaco2005 Mar]Yogurt factory*&&bzoj1740[Usaco2005 mar]Yogurt factory 奶酪工厂*
bzoj1680[Usaco2005 Mar]Yogurt factory bzoj1740[Usaco2005 mar]Yogurt factory 奶酪工厂 题意: n个月,每月有一个酸奶需求量( ...
随机推荐
- Java ConcurrentHashmap 解析
总体描述: concurrentHashmap是为了高并发而实现,内部采用分离锁的设计,有效地避开了热点访问.而对于每个分段,ConcurrentHashmap采用final和内存可见修饰符Volat ...
- IP地址分类及特殊IP地址
A类:0xxx xxxx.x.x.x/8,即1~127,共126个可用. 因0.x.x.x表示所有网络:127.x.x.x/8用作回环地址,作为测试TCP/IP协议的地址. =>其中10.x.x ...
- SQL - 配置SQLServer 使其可以远程访问
环境: SQL Server2008 R2 SQL Server Management Studio 今天测试部署项目的时候,发现不能远程访问SQL Server.具体情形就是在Management ...
- zookeeper[4] 安装windows zookeeper,及问题处理
安装步骤: 1.在如下路径下载zookeeper-3.4.7.tar.gz http://mirrors.cnnic.cn/apache/zookeeper/stable/ 2.解压zookeeper ...
- ELT工具Kettle之CDC(Change Data Capture)实现实例
ETL过程的第一步就是从不同的数据源抽取数据并把数据存储在数据的缓存区.这个过程的主要挑战就是初始加载数据量大和比较慢的网络延迟.在初始加载完成之后,不能再把所有数据重新加载一遍,我们需要的只是变化的 ...
- HTML与CSS简单页面效果实例
本篇博客实现一个HTML与CSS简单页面效果实例 index.html <!DOCTYPE html> <html> <head> <meta charset ...
- Css轮廓
css code: p{ outline-width:2px; outline-color:aqua; outline-style: groove; }
- Ajax请求用户控件(.ascx)404错误
- sublime3 使用技巧
Ctrl+O(Command+O)可以实现头文件和源文件之间的快速切换 Ctrl+Shift+T可以打开之前关闭的tab页,这点同chrome是一样的 Ctrl+R定位函数:Ctrl+G定位到行: 插 ...
- js实现数字分页
///js数字分页 返回数组function page(pageCount, sideNum, pageNum) { //其实页 var startNum = 0; //结束页 ...