【LOJ】#2084. 「NOI2016」网格

题解

之前用的mapTLE了，今天用了个hash把题卡了过去，AC数++

我们只要保留一个点为中心周围5 * 5个格子就可以

如果一个点周围5*5个格子有两个不连通，那么显然输出0

如果一个出现了一个割点，那么看看这个割点在不在离中心点的第一层，如果在的话就是1，没有合法割点的话就是2

然后就是特判了……特判真的挺多的……

代码

#include <bits/stdc++.h>
//#define ivorysi
#define MAXN 100005
#define mo 974711
#define INF 1000000000000000000LL
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
typedef long long int64;
pii poi[MAXN * 26];
vector<int> V[MAXN],st;
int N,M,c,cnt,T;
struct Hash {
    struct node {
        int64 x;int next,c;
    }E[MAXN * 26];
    int head[mo + 5],sumE;
    void clear() {
        sumE = 0;memset(head,0,sizeof(head));
    }
    void add(int64 x,int c) {
        int u = x % mo;
        E[++sumE].x = x;E[sumE].next = head[u];E[sumE].c = c;head[u] = sumE;
    }
    void Insert(pii x,int c) {
        add(1LL * (x.fi - 1) * M + x.se,c);
    }
    int Query(pii x) {
        int u = (1LL * (x.fi - 1) * M + x.se) % mo;
        for(int i = head[u] ; i ; i = E[i].next) {
            if(E[i].x == 1LL * (x.fi - 1) * M + x.se) return E[i].c;
        }
        return -1;
    }
}H;
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
struct node {
    int to,next;
}edge[MAXN * 100];
int head[MAXN * 26],sumE,low[MAXN * 26],dfn[MAXN * 26],idx,col[MAXN * 26],tot;
bool fir[MAXN * 26];
void dfs(int u,int fa) {
    dfn[u] = low[u] = ++idx;
    col[u] = tot;
    int son = 0;
    for(int i = head[u] ; i ; i = edge[i].next) {
	int v = edge[i].to;
	if(v != fa) {
	    if(dfn[v]) low[u] = min(low[u],dfn[v]);
	    else {
		++son;
		dfs(v,u);
		low[u] = min(low[u],low[v]);
		if(fa != 0 && low[v] >= dfn[u]) {
		    st.push_back(u);
		}
	    }
	}
    }
    if(!fa && son > 1) st.push_back(u);
}
void add(int u,int v) {
    edge[++sumE].to = v;
    edge[sumE].next = head[u];
    head[u] = sumE;
}
void Init() {

    sumE = 0;
    for(int i = 1 ; i <= cnt ; ++i) {
	head[i] = dfn[i] = low[i] = col[i] = fir[i] = tot = idx = 0;
    }
    H.clear();
    int u,v;
    for(int i = 1 ; i <= c ; ++i) {
	scanf("%d%d",&u,&v);
	poi[i] = make_pair(u,v);
	H.Insert(poi[i],i);
    }

    cnt = c;
    for(int i = 1 ; i <= c ; ++i) {
	V[i].clear();
	for(int x = -2 ; x <= 2 ; ++x) {
	    for(int y = -2 ; y <= 2 ; ++y) {
		if(x == 0 && y == 0) continue;
		pii tmp = make_pair(poi[i].fi + x,poi[i].se + y);
		if(tmp.fi < 1 || tmp.fi > N || tmp.se < 1 || tmp.se > M) continue;
		if(H.Query(tmp) == -1) {
		    poi[++cnt] = tmp;
		    H.Insert(poi[cnt],cnt);
		}
		if(x <= 1 && x >= -1 && y <= 1 && y >= -1) fir[H.Query(tmp)] = 1;
		V[i].push_back(H.Query(tmp));
	    }
	}
    }
    for(int i = c + 1 ; i <= cnt ; ++i) {
	for(int j = 0 ; j <= 3 ; ++j) {
	    pii tmp = make_pair(poi[i].fi + dx[j],poi[i].se + dy[j]);
        if(tmp.fi < 1 || tmp.fi > N || tmp.se < 1 || tmp.se > M) continue;
	    int x = H.Query(tmp);
	    if(x <= c) continue;
	    add(i,x);
	}
    }
}
void Solve() {
    scanf("%d",&T);
    while(T--) {
	scanf("%d%d%d",&N,&M,&c);
	Init();
	if(1LL * N * M - c <= 1) {puts("-1");goto again;}
	if(c == 0) {
	    if(1LL * N * M == 2) puts("-1");
	    else if(N == 1 || M == 1) puts("1");
	    else puts("2");
	    goto again;
	}
	st.clear();
	for(int i = c + 1 ; i <= cnt ; ++i) {
	    if(!dfn[i]) {
		++tot;
		dfs(i,0);
	    }
	}
	for(int i = 1 ; i <= c ; ++i) {
	    int t = 0;
	    for(auto k : V[i]) {
		if(k <= c) continue;
		if(k > c) {
		    if(t == 0) t = col[k];
		    else if(t != col[k]) {
			puts("0");
			goto again;
		    }
		}
	    }
	}
	if(1LL * N * M - c == 2) {puts("-1");goto again;}
	for(auto k : st) {
	    if(fir[k]) {puts("1");goto again;}
	}
	if(M == 1 || N == 1) puts("1");
	else puts("2");
        again:;
    }
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}