BZOJ1924 [Sdoi2010]所驼门王的宝藏 【建图 + tarjan】

题目

输入格式

第一行给出三个正整数 N, R, C。 以下 N 行，每行给出一扇传送门的信息，包含三个正整数xi, yi, Ti，表示该传送门设在位于第 xi行第yi列的藏宝宫室，类型为 Ti。Ti是一个1~3间的整数， 1表示可以传送到第 xi行任意一列的“横天门”，2表示可以传送到任意一行第 yi列的“纵寰门”，3表示可以传送到周围 8格宫室的“ziyou门”。 保证 1≤xi≤R，1≤yi≤C，所有的传送门位置互不相同。

输出格式

只有一个正整数，表示你确定的路线所经过不同藏宝宫室的最大数目。

输入样例

10 7 7

2 2 1

2 4 2

1 7 2

2 7 3

4 2 2

4 4 1

6 7 3

7 7 1

7 5 2

5 2 1

输出样例

9

提示

题解

把图建出来后就是tarjan + dp了

建图比较烦

我们将其按x排序，先将横向边建上，当一排存在多个横向边时，它们之间两两连边，过于费空间，只连一个环就好了

竖向边也是一样的

最后是第三种边，排个序二分查找就好了

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 100005,maxm = 1000005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int n,N,M;
int x[maxn],y[maxn],id[maxn],t[maxn];
inline bool cmp1(const int& a,const int& b){
	return x[a] == x[b] ? t[a] < t[b] : x[a] < x[b];
}
inline bool cmp2(const int& a,const int& b){
	return y[a] == y[b] ? t[a] < t[b] : y[a] < y[b];
}
inline bool cmp3(const int& a,const int& b){
	return x[a] == x[b] ? y[a] < y[b] : x[a] < x[b];
}
int h[maxn],ne = 2,h2[maxn],ne2 = 2;
struct EDGE{int to,nxt;}ed[maxm],e[maxm];
inline void build(int u,int v){
	ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
	//printf("build (%d)  to   (%d)\n",u,v);
}
inline void add(int u,int v){
	e[ne2] = (EDGE){v,h2[u]}; h2[u] = ne2++;
}
inline bool check(const int& p,const int& X,const int& Y){
	return x[id[p]] == X ? y[id[p]] >= Y : x[id[p]] > X;
}
int lowb(int X,int Y){
	int l = 1,r = n,mid;
	while (l < r){
		mid = l + r >> 1;
		if (check(mid,X,Y)) r = mid;
		else l = mid + 1;
	}
	return l;
}
int st[maxn],top;
void Build(){
	sort(id + 1,id + 1 + n,cmp1);
	for (int i = 1; i <= n; i++){
		int u = i; top = 0;
		while (u <= n){
			if (x[id[u]] != x[id[i]]) break;
			if (t[id[u]] == 1) st[++top] = id[u];
			u++;
		}
		if (!top) continue;
		if (top > 1){
			for (int j = 1; j < top; j++) build(st[j],st[j + 1]);
			build(st[top],st[1]);
		}
		for (int j = i; j < u; j++) if (t[id[j]] != 1) build(st[1],id[j]);
		i = --u;
	}
	sort(id + 1,id + 1 + n,cmp2);
	for (int i = 1; i <= n; i++){
		int u = i; top = 0;
		while (u <= n){
			if (y[id[u]] != y[id[i]]) break;
			if (t[id[u]] == 2) st[++top] = id[u];
			u++;
		}
		if (!top) continue;
		if (top > 1){
			for (int j = 1; j < top; j++) build(st[j],st[j + 1]);
			build(st[top],st[1]);
		}
		for (int j = i; j < u; j++) if (t[id[j]] != 2) build(st[1],id[j]);
		i = --u;
	}
	sort(id + 1,id + 1 + n,cmp3);
	for (int i = 1; i <= n; i++){
		if (t[id[i]] != 3) continue;
		int u = lowb(x[id[i]] - 1,y[id[i]] - 1);
		for (int j = u; j <= n && x[id[j]] + 1 == x[id[i]] && y[id[j]] >= y[id[i]] - 1 && y[id[j]] <= y[id[i]] + 1; j++)
			build(id[i],id[j]);
		u = lowb(x[id[i]],y[id[i]] - 1);
		for (int j = u; j <= n && x[id[j]] == x[id[i]] && y[id[j]] >= y[id[i]] - 1 && y[id[j]] <= y[id[i]] + 1; j++)
			if (id[i] != id[j]) build(id[i],id[j]);
		u = lowb(x[id[i]] + 1,y[id[i]] - 1);
		for (int j = u; j <= n && x[id[j]] - 1 == x[id[i]] && y[id[j]] >= y[id[i]] - 1 && y[id[j]] <= y[id[i]] + 1; j++)
			build(id[i],id[j]);
	}
}
int dfn[maxn],low[maxn],Scc[maxn],scci,cnt,val[maxn];
void dfs(int u){
	dfn[u] = low[u] = ++cnt;
	st[++top] = u;
	Redge(u){
		if (!dfn[to = ed[k].to]){
			dfs(to);
			low[u] = min(low[u],low[to]);
		}else if (!Scc[to]) low[u] = min(low[u],dfn[to]);
	}
	if (dfn[u] == low[u]){
		scci++;
		do{
			Scc[st[top]] = scci; val[scci]++;
		}while(st[top--] != u);
	}
}
void tarjan(){
	for (int i = 1; i <= n; i++) if (!dfn[i]) dfs(i);
}
queue<int> q;
int inde[maxn],f[maxn];
void solve(){
	for (int i = 1; i <= n; i++){
		int u = Scc[i];
		Redge(i) if (Scc[to = ed[k].to] != u)
			add(u,Scc[to]),inde[Scc[to]]++;
	}
	for (int i = 1; i <= scci; i++) if (!inde[i]) q.push(i),f[i] = val[i];
	int u,ans = 0;
	while (!q.empty()){
		u = q.front(); q.pop();
		ans = max(ans,f[u]);
		for (int k = h2[u],to; k; k = e[k].nxt){
			f[to = e[k].to] = max(f[to],f[u] + val[to]);
			if (!(--inde[to])) q.push(to);
		}
	}
	for (int i = 1; i <= scci; i++)
		ans = max(ans,f[i]);
	printf("%d\n",ans);
}
void readin(){
	n = read(); N = read(); M = read();
	for (int i = 1; i <= n; i++)
		x[i] = read(),y[i] = read(),t[i] = read(),id[i] = i;
}
int main(){
	readin();
	Build();
	tarjan();
	solve();
	return 0;
}