UVALive 2664 One-way traffic

One-way traffic

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVALive. Original ID: 2664
64-bit integer IO format: %lld      Java class name: Main

 

In a certain town there are n intersections connected by two- and one-way streets. The town is very modern so a lot of streets run through tunnels or viaducts. Of course it is possible to travel between any two intersections in both ways, i.e. it is possible to travel from an intersection a to an intersection b as well as from b to a without violating traffic rules. Because one-way streets are safer, it has been decided to create as much one-way traffic as possible. In order not to make too much confusion it has also been decided that the direction of traffic in already existing one-way streets should not be changed.

Your job is to create a new traffic system in the town. You have to determine the direction of traffic for as many two-way streets as possible and make sure that it is still possible to travel both ways between any two intersections.

 

Write a program that:

• reads a description of the street system in the town from the standard input,
• for each two-way street determines one direction of traffic or decides that the street must remain two-way,
• writes the answer to the standard output.

Input

The first line of the input contains two integers n and m, where 2n2000 and n - 1mn(n - 1)/2. Integer n is the number of intersections in the town and integer m is the number of streets.

Each of the next m lines contains three integers a, b and c, where 1an, 1bn, ab and c belongs to {1, 2}. If c = 1 then intersections a and b are connected by an one-way street from a to b. If c = 2 then intersections a and b are connected by a two-way street. There is at most one street connecting any two intersections.

Output

The output contains exactly the same number of lines as the number of two-way streets in the input. For each such street (in any order) the program should write three integers a, b and c meaning, the new direction of the street from a to b (c = 1) or that the street connecting a and b remains two-way (c = 2). If there are more than one solution with maximal number of one-way streets then your program should output any of them but just one.

 

Sample Input

4 4
4 1 1
4 2 2
1 2 1
1 3 2

Sample Output

2 4 1
3 1 2

Source

Regionals 2002, Europe - Central

 

解题：将混合图中的无向边重定向使其成为强连通

 

好吧 解释下 为什么遇到桥，要反向吧！。。。因为不反向，就不能返祖，这样就无法强联通了。。。强连通必须保证low[u]<=dfn[u].

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2020;
struct arc{
    int to,next;
    bool cut,print,flag;
}e[maxn*maxn];
int head[maxn],tot,n,m;
void add(int u,int v,int flag){
    e[tot].to = v;
    e[tot].flag = flag;
    e[tot].cut = false;
    e[tot].print = flag == 2;
    e[tot].next = head[u];
    head[u] = tot++;
}
int low[maxn],dfn[maxn],idx;
void tarjan(int u,int fa){
    bool flag = false;
    low[u] = dfn[u] = ++idx;
    for(int i = head[u]; ~i; i = e[i].next){
        if(e[i].to == fa && !flag){
            flag = true;
            continue;
        }
        if(!dfn[e[i].to]){
            tarjan(e[i].to,u);
            low[u] = min(low[u],low[e[i].to]);
            if(low[e[i].to] > dfn[u]) e[i].cut = e[i^1].cut = true;
        }else low[u] = min(low[u],dfn[e[i].to]);
    }
}
void dfs(int u,int fa){
    low[u] = dfn[u] = ++idx;
    for(int i = head[u]; ~i; i = e[i].next){
        if(e[i].to == fa || !e[i].flag) continue;
        e[i].flag = true;
        e[i^1].flag = false;
        if(!dfn[e[i].to]){
            dfs(e[i].to,u);
            low[u] = min(low[u],low[e[i].to]);
            if(low[e[i].to] > dfn[u]){
                e[i].flag = false;
                e[i^1].flag = true;
            }
        } else low[u] = min(low[u],dfn[e[i].to]);
    }
}
int main(){
    int u,v,t;
    while(~scanf("%d%d",&n,&m)){
        memset(head,-1,sizeof head);
        for(int i = tot = 0; i < m; ++i){
            scanf("%d%d%d",&u,&v,&t);
            if(t == 2){
                add(u,v,2);
                add(v,u,2);
            }else{
                add(u,v,1);
                add(v,u,0);
            }
        }
        idx = 0;
        memset(dfn,0,sizeof dfn);
        tarjan(1,-1);
        idx = 0;
        memset(dfn,0,sizeof dfn);
        dfs(1,-1);
        for(int i = 0; i < tot; i += 2)
        if(e[i].print){
            if(e[i].cut) printf("%d %d 2\n",e[i^1].to,e[i].to);
            else if(e[i].flag)  printf("%d %d 1\n",e[i^1].to,e[i].to);
            else printf("%d %d 1\n",e[i].to,e[i^1].to);
        }
    }
    return 0;
}