題目描述
Now there are n cities and m unidirectional roads, which means if there is a road from a to b, then you can only travel from a to b but never b to a. At first, no two cities can reach each other(means if you travel from a to b, you have no way to get back a, neither direct nor indirect). Now if you have a chance to change an unidirectional road into a bidirectional road, how many pairs of cities will become mutually connected at most? Note that (a,a) is not a legal pair, and (a,b),(b,a) are regarded as one pair.
輸入格式
The first line is case number T(1≤T≤3), T test
cases follow.
For each test case:
- The first line are two integers n,m(1≤n≤1000,1≤m≤5000)
- Then m lines follow, each line contains two integers a,b, which means there is an unidirectional road form city a to city b(1≤a,b≤n).
輸出格式
For each test case print an integer indicating most mutually connected city pairs after the change.
思路還是比較容易想到,原圖是一個有向無環圖,找出該成無向邊後能形成的最大的強聯通分量,上面所有的點都是強聯通的,取C(2,n)就能得到答案。
最開始用BFS瞎搞,然後TLE,向大佬請教思路後方才發現m的大小完全可以用tarjan縮點,m*(m+n)的複雜度。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
#define N 1001
#define M 5005
struct Edge
{
int s;
int v;
int next;
};
Edge edge[M];//邊的集合
int node[N];//頂點集合
int instack[N];//標記是否在stack中
int stack[N];
int Belong[N];//各頂點屬於哪個強連通分量
int DFN[N];//節點u搜索的序號(時間戳)
int LOW[N];//u或u的子樹能夠追溯到的最早的棧中節點的序號(時間戳)
int n, m;//n:點的個數;m:邊的條數
int cnt_edge;//邊的計數器
int Index;//序號(時間戳)
int top;
int Bcnt;//有多少個強連通分量
int num[N];
int ans;
int dis(int k){
return k*(k-1)/2;
}
void add_edge(int u, int v)//鄰接表存儲
{
edge[cnt_edge].s = u;
edge[cnt_edge].next = node[u];
edge[cnt_edge].v = v;
node[u] = cnt_edge++;
}
void tarjan(int u)
{
int i,j;
int v;
DFN[u]=LOW[u]=++Index;
instack[u]=true;
stack[++top]=u;
for (i = node[u]; i != -1; i = edge[i].next)
{
v=edge[i].v;
if (!DFN[v])//如果點v沒被訪問
{
tarjan(v);
if (LOW[v]<LOW[u])
LOW[u]=LOW[v];
}
else//如果點v已經被訪問過
if (instack[v] && DFN[v]<LOW[u])
LOW[u]=DFN[v];
}
if (DFN[u]==LOW[u])
{
Bcnt++;
do
{
j=stack[top--];
instack[j]=false;
Belong[j]=Bcnt;
}
while (j!=u);
}
}
void solve()
{
int i;
top=Bcnt=Index=0;
memset(DFN,0,sizeof(DFN));
memset(LOW,0,sizeof(LOW));
for (i=1;i<=n;i++)
if (!DFN[i])
tarjan(i);
}
int main()
{
int T;
int i,j,k;
int sum,temp;
scanf("%d",&T);
while(T--){
ans = 0;
cnt_edge = 0;
memset(node,-1,sizeof(node));
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++)
{
scanf("%d%d",&j,&k);
add_edge(j,k);
}
for(int p=0;p<m;p++){
sum =0;
memset(num,0,sizeof(num));
edge[cnt_edge].s = edge[p].v;
edge[cnt_edge].next = node[edge[p].v];
edge[cnt_edge].v = edge[p].s;
temp = node[edge[p].v];
node[edge[p].v] = cnt_edge++;
solve();
cnt_edge--;
node[edge[p].v] = temp;
for(i=1;i<=n;i++){
num[Belong[i]]++;
}
for(i=1;i<=n;i++){
sum = max(sum,num[i]);
}
ans = max(ans,sum);
//printf("no.%d. ans: %d\n",p,ans);
}
printf("%d\n",dis(ans));
}
return 0;
}