周赛 HDU 2767 1269 1872 强连通 - c++编程基础

add(a,b) ;
}
for (int i = 1 ;i <= n; i ++)if(dfn[i] == -1)tarjan(i) ;
//cout < if(num == 1)cout <<"Yes"< else
cout <<"No"< }
}

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#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
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 < if(num == 1)cout <<"Yes"< else
cout <<"No"< }
}

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。

[cpp]
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#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
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] ;vectorqq[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

周赛 HDU 2767 1269 1872 强连通 (三)