周赛 HDU 2767 1269 1872 强连通
HDU 2767
题意：给出一些点之间的关系，然后问最少添加多少条边可以使这张图强连通。
裸题，唯一的trick就是判断图一开始就是强连通图的时候输出为0，这里没想清楚，导致卡了半小时。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 100005
#define inf 1<<28
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define FOR(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
using namespace std;
struct kdq
{
int e ,next ;
} ed[Max * 10] ;
int head[Max] ,num1 ;
int dfn[Max] ,low[Max] , vis[Max] , st[Max] , in[Max] ,out[Max] ,belong[Max] ,cnt[Max] ;
int tp = 0 ,dp = 0 ,num = 0 ;
void add(int a ,int b)
{
ed[num1].e = b ;
ed[num1].next = head[a] ;
head[a] = num1 ++ ;
}
void init()
{
mem(dfn,-1) ;
mem(low,0) ;
mem(vis,0) ;
mem(st,0) ;
mem(in,0) ;
mem(out,0) ;
mem(belong,0) ;
mem(head,-1) ;
mem(cnt,0) ;
num = num1 = tp = dp = 0 ;
}
void tarjan(int now)
{
vis[now] = 1 ;
st[tp ++ ] = now ;
dfn[now] = low[now] = dp ++ ;
for (int i = head[now] ; i != -1 ; i = ed[i].next )
{
int v = ed[i].e ;
if(dfn[v] == -1)
{
tarjan(v) ;
low[now] = min(low[now] ,low[v]) ;
}
else if(vis[v])
{
low[now] = min(low[now] ,dfn[v]) ;
}
}
if(low[now] == dfn[now])
{
int xx ;
num ++ ;
do
{
xx = st[-- tp ] ;
vis[xx] = 0 ;
belong[xx] = num ;
cnt[num] ++ ;
}
while(xx != now) ;
}
}
void solve()
{
int n ,m ;
int T ;
while(cin >> T )
{
while( T -- )
{
cin >> n>> m ;
init() ;
for (int i = 0 ; i < m ; i ++)
{
int a , b ;
scanf("%d%d",&a,&b) ;
add(a,b) ;
}
for (int i = 1 ; i <= n ; i ++ )
if(dfn[i] == -1)tarjan(i) ;
for (int i = 1 ; i <= n ; i ++ )
for (int j = head[i] ; j != -1 ; j = ed[j].next )
{
int x = belong[i] ;
int y = belong[ed[j].e] ;
if(x != y )
{
out[x] ++ ;
in[y] ++ ;
}
}
int ans = 0 ;
int ans1 = 0 ;
int kk = 0 ;
mem(vis,0) ;
for (int i = 1 ; i <= num ; i ++)
{
if(in[i] == 0)ans ++ ;
if(out[i] == 0)ans1 ++ ;
}
ans1 = max(ans,ans1) ;
//ans1 += ans ;
if(num == 1)ans1 = 0 ;
cout <<ans1 <<endl;
}
}
}
int main()
{
solve() ;
return 0;
}
HDU 1269
裸题
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 100005
#define inf 1<<28
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define FOR(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define PII pair<int,int>
using namespace std;
struct kdq
{
int e, l, next ;
}ed[Max] ;
int head[Max] ;
int num1 ;
void add(int a,int b)
{
ed[num1].e = b ;
ed[num1].next = head[a] ;
head[a] = num1 ++ ;
}
int dfn[Max] ,low[Max] ,st[Max] ,vis[Max] ;
int belong[Max] ;
int dp , tp ,num ;
void init()
{
mem(head,-1) ;
num1 = 0 ;
mem(dfn,-1) ;
mem(low,0) ;
mem(st,0) ;
mem(vis,0) ;
mem(belong ,0) ;
dp = 0 ;
tp = 0 ;
num = 0 ;
}
void tarjan(int now)
{
dfn[now] = low[now] = dp ++ ;
vis[now] = 1 ;
st[tp ++ ] = now ;
for (int i = head[now] ; i != -1 ;i = ed[i].next )
{
int e = ed[i].e ;
if(dfn[e] == -1)
{tarjan(e);
low[now] = min(low[now],low[e]) ;
}
else if(vis[e])
low[now] = min(low[now],dfn[e]) ;
}
if(dfn[now] == low[now])
{
num ++ ;
int xx ;
do
{
xx = st[-- tp] ;
vis[xx] = 0 ;
belong[xx] = num ;
}
while(xx != now) ;
}
}
int main()
{
int n ,m ;
while(scanf("%d%d",&n,&m),(n + m))
{
init() ;
for (int i = 0 ;i < m ;i ++)
{
int a ,b ;
scanf("%d%d",&a,&b) ;
add(a,b) ;
}
for (int i = 1 ;i <= n; i ++)if(dfn[i] == -1)tarjan(i) ;
//cout <<num<<endl;
if(num == 1)cout <<"Yes"<<endl;
else
cout <<"No"<<endl;
}
}
HDU 1872
题意：中文题，不解释。
思路：tarjan，缩点，然后找出入度为0的联通分量，然后求出这些分量里花费最少的一个人，相加即可。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 100005
#define inf 1<<28
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define FOR(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define PII pair<int,int>
using namespace std;
struct kdq
{
int e, l, next ;
} ed[Max] ;
int cost[Max] ;
int head[Max] ;
int num1 ;
void add(int a,int b)
{
ed[num1].e = b ;
ed[num1].next = head[a] ;
head[a] = num1 ++ ;
}
int dfn[Max] ,low[Max] ,st[Max] ,vis[Max] ;
int belong[Max] ;
int dp , tp ,num ;
int in[Max] ,out[Max] ;vector<int>qq[Max] ;
void init(int n)
{
mem(head,-1) ;
num1 = 0 ;
mem(dfn,-1) ;
mem(low,0) ;
mem(st,0) ;
mem(vis,0) ;
mem(belong ,0) ;
dp = 0 ;
tp = 0 ;
num = 0 ;
mem(in,0) ;
mem(out,0) ;
for (int i = 0 ;i <= n ;i ++)qq[i].clear() ;
}
void tarjan(int now)
{
dfn[now] = low[now] = dp ++ ;
vis[now] = 1 ;
st[tp ++ ] = now ;
for (int i = head[now] ; i != -1 ; i = ed[i].next )
{
int e = ed[i].e ;
if(dfn[e] == -1)
{
tarjan(e);
low[now] = min(low[now],low[e]) ;
}
else if(vis[e])
low[now] = min(low[now],dfn[e]) ;
}
if(dfn[now] == low[now])
{
num ++ ;
int xx ;
do
{
xx = st[-- tp] ;
vis[xx] = 0 ;
qq[num].push_back(xx) ;
belong[xx] =num ;
}
while(xx != now) ;
}
}
int main()
{
int n ,m ;
while(scanf("%d%d",&n,&m) != EOF)
{
init(n) ;
for (int i = 1 ; i <= n ; i ++)scanf("%d",&cost[i]) ;
for (int i = 0 ; i < m ; i ++)
{
int a ,b ;
scanf("%d%d",&a,&b) ;
add(a,b) ;
}
for (int i = 1 ; i <= n; i ++)if(dfn[i] == -1)tarjan(i) ;
for (int i = 1 ; i <= n ; i ++)
{
for (int j = head[i] ; j != -1 ; j = ed[j].next )
{
int x = belong[i] ;
int y = belong[ed[j].e] ;
if(x != y )
{
out[x] ++ ;
in[y] ++ ;
}
}
}
ll aa = 0 ;
int cc = 0 ;
int sum = inf ;
int i ;
for (i = 1 ; i <= num ; i ++)
{
if(in[i] == 0)
{
cc ++ ;
sum = inf ;
int l = qq[i].size() ;
for (int k = 0 ; k < l ; k ++)
sum = min(sum,cost[qq[i][k]]) ;
aa += sum ;
}
}
cout <<cc <<" "<<aa <<endl;
}
}