2017-2018 ACM-ICPC Southeast Regional Contest (Div. 1) H.Security Badges 区间双重合

http://codeforces.com/gym/101617


题意:这个老板想检查自己酒店的安全性,他的手下有一些保安,每个保安有自己一个特有的带有标号的勋章,房间与房间之前有门相连。

每个门只允许特定的一个范围的勋章标号打开门。问:有多少个保安能从s走到d。


首先对于这个题能够建图,并且我们清楚不断的从一个房间进入下一个房间,勋章范围一定是在减小或者不变。

那么最后的范围一定是所有房间允许范围的交集,那么我们假如记录房间允许范围所有的点,然后排序呢?

这样每相邻的两点,一定是最小的区间,假如这个区间范围的点能从s走到d的话,那么这个区间就算在答案里面。

那么思路定下来之后,就得跑一个dfs了。

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair  ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
templateinline T min(T a,T b,T c) { return min(min(a,b),c);}
templateinline T max(T a,T b,T c) { return max(max(a,b),c);}
templateinline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
templateinline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template 
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;


int n,m,k,s,d;
struct node 
{
	int to,a,b;
};
vectorG[maxn];
vectorg;
bool vis[maxn];
int dfs(int u,int a,int b)
{
	vis[u]=1;
	if(u==d) return 1;
	for(auto v:G[u])
	{
		if(vis[v.to]) continue;
		if(v.a<=a&&v.b>=b)
			if(dfs(v.to,a,b)) return 1;
	}
	return 0;
}
		
void solve()
{
	s_3(n,m,k);
	s_2(s,d);
	FOR(1,m,i)
	{
		int u,v,a,b;
		s_4(u,v,a,b);
		G[u].pb({v,a-1,b});
		g.pb(a-1);
		g.pb(b);
	}
	sort(g.begin(),g.end());
	int pri=0,ans=0;
	for(auto pto:g)
	{
		me(vis,0);
		ans+=(pto-pri)*dfs(s,pri,pto);
		pri=pto;
	}
	print(ans);
}
int main()
{
    //freopen( "1.in" , "r" , stdin );
    //freopen( "1.out" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++)
    {
        //printf("Case #%d: ",cas);
        solve();
    }
}


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