题意
一个(r imes c)的棋盘,棋盘上有(n)个标记点,每个点有三种类型,类型(1)可以传送到本行任意标记点,类型(2)可以传送到本列任意标记点,类型(3)可以传送到周围八连通任意标记点。求最长路径。
(r,cleq 10^6,nleq 10^5)
题解
这题做法很多,我就把每一行的所有类型(1)门缩到一起(直接找一个代表),列也同理,然后暴力连边,类型(3)连边用( ext{map}),这样每个点的入边中类型(1)或(2)最多有(1)条,类型(3)最多(8)条,大概可以说明边数和点数同阶,于是( ext{Tarjan})缩点然后(dp)求最长路...
#include <algorithm>
#include <utility>
#include <cstdio>
#include <vector>
#include <stack>
#include <map>
using namespace std;
const int N = 1e5 + 10;
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
struct node {
int x, y, z, sz;
} a[N];
int n, r, c, f[N], rt[2][N * 10], dT[N];
vector<int> ob[2][N * 10], G[N], T[N];
map<pair<int, int>, int> ma;
bool isr[N];
int dfn[N], low[N], sz[N], bel[N], scc;
stack<int> st;
bool ins[N];
void tarjan(int u) {
low[u] = dfn[u] = ++ dfn[0];
st.push(u); ins[u] = 1;
for(int i = 0; i < G[u].size(); i ++) {
int v = G[u][i];
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if(ins[v]) {
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u]) {
scc ++;
while(1) {
int v = st.top(); st.pop();
ins[v] = 0; bel[v] = scc;
sz[scc] += a[v].sz;
if(u == v) break ;
}
}
}
int pa[N];
int solve(int u) {
if(pa[u]) return pa[u];
for(int i = 0; i < T[u].size(); i ++) {
pa[u] = max(pa[u], solve(T[u][i]));
}
pa[u] += sz[u];
return pa[u];
}
int main() {
scanf("%d%d%d", &n, &r, &c);
for(int i = 1; i <= n; i ++) {
scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);
ma[make_pair(a[i].x, a[i].y)] = i;
ob[0][a[i].x].push_back(i);
ob[1][a[i].y].push_back(i);
a[i].sz = 0;
}
for(int i = 1; i <= n; i ++) {
if(a[i].z == 1) {
int &u = rt[0][a[i].x];
if(!u) u = i;
a[u].sz ++; f[i] = u;
}
if(a[i].z == 2) {
int &u = rt[1][a[i].y];
if(!u) u = i;
a[u].sz ++; f[i] = u;
}
if(a[i].z == 3) {
f[i] = i;
a[i].sz ++;
}
isr[f[i]] = 1;
}
for(int i = 1; i <= n; i ++)
if(isr[i]) {
if(a[i].z == 1) {
for(int j = 0; j < ob[0][a[i].x].size(); j ++) {
int v = ob[0][a[i].x][j];
G[i].push_back(f[v]);
}
}
if(a[i].z == 2) {
for(int j = 0; j < ob[1][a[i].y].size(); j ++) {
int v = ob[1][a[i].y][j];
G[i].push_back(f[v]);
}
}
if(a[i].z == 3) {
for(int j = 0; j < 8; j ++) {
int v = ma[make_pair(a[i].x + dx[j], a[i].y + dy[j])];
if(v) {
G[i].push_back(f[v]);
}
}
}
}
for(int i = 1; i <= n; i ++)
if(isr[i] && !dfn[i]) {
tarjan(i);
}
for(int i = 1; i <= n; i ++) {
if(isr[i]) {
for(int j = 0; j < G[i].size(); j ++) {
int v = G[i][j];
if(bel[i] != bel[v]) {
T[bel[i]].push_back(bel[v]);
dT[bel[v]] ++;
}
}
}
}
int ans = 0;
for(int i = 1; i <= scc; i ++)
if(!dT[i]) {
ans = max(ans, solve(i));
}
printf("%d
", ans);
return 0;
}