2 seconds
256 megabytes
standard input
standard output
Andrew prefers taxi to other means of transport, but recently most taxi drivers have been acting inappropriately. In order to earn more money, taxi drivers started to drive in circles. Roads in Andrew's city are one-way, and people are not necessary able to travel from one part to another, but it pales in comparison to insidious taxi drivers.
The mayor of the city decided to change the direction of certain roads so that the taxi drivers wouldn't be able to increase the cost of the trip endlessly. More formally, if the taxi driver is on a certain crossroads, they wouldn't be able to reach it again if he performs a nonzero trip.
Traffic controllers are needed in order to change the direction the road goes. For every road it is known how many traffic controllers are needed to change the direction of the road to the opposite one. It is allowed to change the directions of roads one by one, meaning that each traffic controller can participate in reversing two or more roads.
You need to calculate the minimum number of traffic controllers that you need to hire to perform the task and the list of the roads that need to be reversed.
The first line contains two integers nn and mm (2≤n≤1000002≤n≤100000, 1≤m≤1000001≤m≤100000) — the number of crossroads and the number of roads in the city, respectively.
Each of the following mm lines contain three integers uiui, vivi and cici (1≤ui,vi≤n1≤ui,vi≤n, 1≤ci≤1091≤ci≤109, ui≠viui≠vi) — the crossroads the road starts at, the crossroads the road ends at and the number of traffic controllers required to reverse this road.
In the first line output two integers the minimal amount of traffic controllers required to complete the task and amount of roads kk which should be reversed. kk should not be minimized.
In the next line output kk integers separated by spaces — numbers of roads, the directions of which should be reversed. The roads are numerated from 11 in the order they are written in the input. If there are many solutions, print any of them.
5 6
2 1 1
5 2 6
2 3 2
3 4 3
4 5 5
1 5 4
2 2
1 3
5 7
2 1 5
3 2 3
1 3 3
2 4 1
4 3 5
5 4 1
1 5 3
3 3
3 4 7
There are two simple cycles in the first example: 1→5→2→11→5→2→1 and 2→3→4→5→22→3→4→5→2. One traffic controller can only reverse the road 2→12→1 and he can't destroy the second cycle by himself. Two traffic controllers can reverse roads 2→12→1 and 2→32→3 which would satisfy the condition.
In the second example one traffic controller can't destroy the cycle 1→3→2→11→3→2→1. With the help of three controllers we can, for example, reverse roads 1→31→3 ,2→42→4, 1→51→5.
【题意】
给定一张有向图，每条边有边权。你可以花费边权的代价反转一条边，使得原图中没有环。
1、输出最小化的反转的边权的最大值和要反转几条边k（k不必为最小数量）
2、输出你要反转的k条边的序号。（在满足最小化最大值的前提下，任何一种方案皆可）
【分析】
转化为有些边可以翻转，有些边不可以翻转，使得图中没有环。由此二分答案
我们把不能反向的边拿出来，然后跑拓扑排序判环，如果有环则无解，不然一定有一种方案，加入那些可以改变方向的边而不产生环。
新加的边方向：拓扑序小的连向拓扑序大的
Attached official solution.
Suppose we have k traffic controllers. They can turn all edges whose weight is less than or equal to k. Then let's remove all these edges from the graph, make a topological sorting of the remaining graph, and orient the other edges in the order of topological sorting. If there are cycles left in the graph after removing the edges, then we cannot get rid of them, having k traffic controllers. Otherwise, by adding edges we will not add new loops. The parameter k can be iterated through a binary search. Also in binary search, you can go through not all possible values of k, but only the values that are on the edges.
Complexity — O((n+m)logC) or O((n+m)logm).
【代码】
#include<stack>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<'\n';
using namespace std;
inline int read(){
register char ch=getchar();register int x=0;
for(;ch<'0'||ch>'9';ch=getchar());
for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
return x;
}
const int N=1e5+5;
int n,m,cct,in[N],dfn[N],rec[N];
struct data{
int x,y,z;
data(int _x=0,int _y=0,int _z=0){
x=_x;y=_y;z=_z;
}
}b[N];
struct edge{int v,next;}e[N];int tot,head[N];
inline void add(int x,int y){
e[++tot].v=y;e[tot].next=head[x];head[x]=tot;
}
inline bool topo(){
stack<int>s;int cnt=0;
for(int i=1;i<=n;i++) if(!in[i]) s.push(i),dfn[i]=++cnt;
while(!s.empty()){
int x=s.top();s.pop();
for(int j=head[x];j;j=e[j].next){
int v=e[j].v;
if(!--in[v]) s.push(v),dfn[v]=++cnt;
}
}
return cnt==n;
}
#define m(a) memset(a,0,(sizeof a[0])*(n+1));
inline bool check(int now){
m(in);m(dfn);m(head);tot=0;
for(int i=1;i<=m;i++)
if(b[i].z>now)
add(b[i].x,b[i].y),in[b[i].y]++;
return topo();
}
int main(){
n=read();m=read();
int l=0,r=0,mid,ans=0;
for(int i=1,x,y,z;i<=m;i++) x=read(),y=read(),z=read(),b[i]=data(x,y,z),r=max(r,z);
while(l<=r){
mid=l+r>>1;
if(check(mid)){
ans=mid;
r=mid-1;
}
else{
l=mid+1;
}
}
printf("%d ",ans);
check(ans);
for(int i=1;i<=m;i++){
if(dfn[b[i].x]>dfn[b[i].y]){
rec[++cct]=i;
}
}
printf("%d\n",cct);
for(int i=1;i<=cct;i++) printf("%d ",rec[i]);
return 0;
}