【CodeForces 567E】President and Roads(最短路)
Description
Berland has n cities, the capital is located in city s, and the historic home town of the President is in city t (s ≠ t). The cities are connected by one-way roads, the travel time for each of the road is a positive integer.
Once a year the President visited his historic home town t, for which his motorcade passes along some path from s to t (he always returns on a personal plane). Since the president is a very busy man, he always chooses the path from s to t, along which he will travel the fastest.
The ministry of Roads and Railways wants to learn for each of the road: whether the President will definitely pass through it during his travels, and if not, whether it is possible to repair it so that it would definitely be included in the shortest path from the capital to the historic home town of the President. Obviously, the road can not be repaired so that the travel time on it was less than one. The ministry of Berland, like any other, is interested in maintaining the budget, so it wants to know the minimum cost of repairing the road. Also, it is very fond of accuracy, so it repairs the roads so that the travel time on them is always a positive integer.
Input
The first lines contain four integers n, m, s and t (2 ≤ n ≤ 105; 1 ≤ m ≤ 105; 1 ≤ s, t ≤ n) — the number of cities and roads in Berland, the numbers of the capital and of the Presidents' home town (s ≠ t).
Next m lines contain the roads. Each road is given as a group of three integers ai, bi, li (1 ≤ ai, bi ≤ n; ai ≠ bi; 1 ≤ li ≤ 106) — the cities that are connected by the i-th road and the time needed to ride along it. The road is directed from city ai to city bi.
The cities are numbered from 1 to n. Each pair of cities can have multiple roads between them. It is guaranteed that there is a path from sto t along the roads.
Output
Print m lines. The i-th line should contain information about the i-th road (the roads are numbered in the order of appearance in the input).
If the president will definitely ride along it during his travels, the line must contain a single word "YES" (without the quotes).
Otherwise, if the i-th road can be repaired so that the travel time on it remains positive and then president will definitely ride along it, print space-separated word "CAN" (without the quotes), and the minimum cost of repairing.
If we can't make the road be such that president will definitely ride along it, print "NO" (without the quotes).
6 7 1 6
1 2 2
1 3 10
2 3 7
2 4 8
3 5 3
4 5 2
5 6 1
YES
CAN 2
CAN 1
CAN 1
CAN 1
CAN 1
YES
3 3 1 3
1 2 10
2 3 10
1 3 100
YES
YES
CAN 81
2 2 1 2
1 2 1
1 2 2
YES
NO
The cost of repairing the road is the difference between the time needed to ride along it before and after the repairing.
In the first sample president initially may choose one of the two following ways for a ride:1 → 2 → 4 → 5 → 6 or 1 → 2 → 3 → 5 → 6.
dijkstra找出从s出发的最短路和从t出发的最短路。
存边的时候正向和逆向分别存起来,并且在求最短路的同时计算到每个点的最短路数量。
d[0][i]表示s出发到i的最短路,d[1][i]表示t出发到i的最短路。每条边的权值为w。
则当w+d[0][u]+d[1][v]时说明是s到t的最短路上的边,如果是所有最短路都经过的边,则满足path[0][u]*path[1][v]==path[0][t]。
path是最短路的数量,因为可能爆long long,因此要取模,而且还不能是1000000007。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define N 100005
#define M 5462617
#define inf 0x3f3f3f3f3f3f3f3fll
#define ll long long
#define add(u,v,w) e[++cnt]=(edge){v,head[0][u],w};head[0][u]=cnt;e[++cnt]=(edge){u,head[1][v],w};head[1][v]=cnt
using namespace std;
struct edge{
ll to,next,w;
}e[N<<];
struct road{
ll u,v,w;
}l[N];
struct qnode{
ll v,c;
bool operator <(const qnode &r)const{
return c>r.c;
}
};
ll n,m,s,t,u,v,w,d[][N],cnt,head[][N],b[][N],path[][N];
void dijkstra(int f){
int i,j,k,pr=f?t:s;
for(i=;i<=n;i++)d[f][i]=inf;
priority_queue<qnode> q;
d[f][pr]=;
q.push((qnode){pr,});
path[f][pr]=;
while(!q.empty()){
qnode u=q.top();
q.pop();
if(b[f][u.v])continue;
b[f][pr=u.v]=;
for(j=head[f][pr];j;j=e[j].next){
k=e[j].to;
if(d[f][pr]+e[j].w==d[f][k])
path[f][k]=(path[f][k]+path[f][pr])%M;
else if(d[f][pr]+e[j].w<d[f][k]){
d[f][k]=d[f][pr]+e[j].w;
q.push((qnode){k,d[f][k]});
path[f][k]=path[f][pr];
}
}
}
}
int main() {
int i,j;
cin>>n>>m>>s>>t;
for(i=;i<=m;i++){
scanf("%lld%lld%lld",&u,&v,&w);
add(u,v,w);
l[i]=(road){u,v,w};
}
dijkstra();
dijkstra();
for(i=;i<=m;i++)
{
u=l[i].u;
v=l[i].v;
w=l[i].w;
if(d[][u]+d[][v]+w==d[][t]){
if(path[][u]*path[][v]%M==path[][t])
puts("YES");
else if(w>)
puts("CAN 1");
else puts("NO");
}
else if(d[][u]+d[][v]+<d[][t])
printf("CAN %lld\n",d[][u]+d[][v]+w-d[][t]+);
else
puts("NO");
}
}
【CodeForces 567E】President and Roads(最短路)的更多相关文章
- Codeforces.567E.President and Roads(最短路 Dijkstra)
题目链接 \(Description\) 给定一张有向图,求哪些边一定在最短路上.对于不一定在最短路上的边,输出最少需要将其边权改变多少,才能使其一定在最短路上(边权必须为正,若仍不行输出NO). \ ...
- Codeforces Round #Pi (Div. 2) E. President and Roads 最短路+桥
题目链接: http://codeforces.com/contest/567/problem/E 题意: 给你一个带重边的图,求三类边: 在最短路构成的DAG图中,哪些边是必须经过的: 其他的(包括 ...
- Codeforces Round #Pi (Div. 2) 567E President and Roads ( dfs and similar, graphs, hashing, shortest paths )
图给得很良心,一个s到t的有向图,权值至少为1,求出最短路,如果是一定经过的边,输出"YES",如果可以通过修改权值,保证一定经过这条边,输出"CAN",并且输 ...
- Codeforces Gym 100338C Important Roads 最短路+Tarjan找桥
原题链接:http://codeforces.com/gym/100338/attachments/download/2136/20062007-winter-petrozavodsk-camp-an ...
- cf567E. President and Roads(最短路计数)
题意 题目链接 给出一张有向图,以及起点终点,判断每条边的状态: 是否一定在最短路上,是的话输出'YES' 如果不在最短路上,最少减去多少权值会使其在最短路上,如果减去后的权值\(< 1\),输 ...
- Codeforces Round #Pi (Div. 2) E. President and Roads tarjan+最短路
E. President and RoadsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/567 ...
- Codeforces 806 D.Prishable Roads
Codeforces 806 D.Prishable Roads 题目大意:给出一张完全图,你需要选取其中的一些有向边,连成一个树形图,树形图中每个点的贡献是其到根节点路径上每一条边的边权最小值,现在 ...
- Codeforces 191C Fools and Roads(树链拆分)
题目链接:Codeforces 191C Fools and Roads 题目大意:给定一个N节点的数.然后有M次操作,每次从u移动到v.问说每条边被移动过的次数. 解题思路:树链剖分维护边,用一个数 ...
- codeforces 689 Mike and Shortcuts(最短路)
codeforces 689 Mike and Shortcuts(最短路) 原题 任意两点的距离是序号差,那么相邻点之间建边即可,同时加上题目提供的边 跑一遍dijkstra可得1点到每个点的最短路 ...
随机推荐
- [No000035]操作系统Operating System之OS Interface操作系统接口
接口(Interface) 仍然从常识开始… 日常生活中有很多接口:电源插座:汽车油门… 那什么是接口? 连接两个东西.信号转换.屏蔽细节… Interface: electrical circuit ...
- uwsgi+flask环境中安装matplotlib
uwsgi+flask的python有自身的virtual environment,可以通过如下命令进入 . venv/bin/activate 虽然通过sudo apt-get install py ...
- iOS 中 const static extern 关键字总结
在看一些高手所写的代码时,总是可以看到我们小白平常不用的关键字,一次,两次,三次,不能总是不明不白,现在总结一下日常开发中常用的关键字的作用: 关键字const/static/extern的释义和用法 ...
- ASP.NET MVC图片上传前预览
回老家过春节,大半个月,在家的日子里,吃好睡好,人也长了3.5Kg.没有电脑,没有网络,无需写代码,工作上相关的完全放下......开心与父母妻儿过个年,那样的生活令Insus.NET现在还在留恋.. ...
- 使用JavaScript打印网页指定DIV区域
JavaScript打印函数myPrint(obj): JavaScript打印页面指定div区域原理:使用window.open()在浏览器打开一个新的页面(window), 使用 window.d ...
- 如何在前台脚本通过json传递数据到后台(使用微软自带的ajax)
首先,我们要在前台引入json的脚本,以便于把js对象序列化 <script type="text/javascript" src="/js/jquery.json ...
- github上传
创建全局的name和email 1.创建ssh(使用命令)$ssh-keygen -t rsa -C xxxxx@gmail.com(注册github时的email)2.在github中添加ssh 登 ...
- C#开发Windows服务
Microsoft Windows 服务(即,以前的 NT 服务)使您能够创建在它们自己的 Windows 会话中可长时间运行的可执行应用程序. 服务可以在计算机启动时自动启动,可以暂停和重新启动而且 ...
- 样条函数 -- spline function
一类分段(片)光滑.并且在各段交接处也有一定光滑性的函数.简称样条.样条一词来源于工程绘图人员为了将一些指定点连接成一条光顺曲线所使用的工具,即富有弹性的细木条或薄钢条.由这样的样条形成的曲线在连接点 ...
- 【Python】[函数] 函数的参数与递归函数
一.函数的参数1.位置参数2.默认参数 n就是默认参数 def power(x,n=2): s=1 while n > 0: n = n - 1 s = s * x return s 默认参数有 ...