BZOJ 1453 [WC] 双面棋盘 并查集+线段树暴搞

这道题大部分这么做的题解写的都很简洁，不过看代码就能看懂大概搞法。

线段树的一个节点维护的信息有：节点对应区间[l,r]为原图第l行到第r行这(r-l+1)×n的棋盘中的黑色连通块数和白色连通块数。然后我们保存第l行和第r行的连通的情况，这样就可以方便地合并两个相邻区间了。注意到同一行中不相邻的两个块可能是属于一个连通块的，所以用并查集来储存连通的信息比较合适。每次合并都是O(n)。合并区间时，连通块数容易维护，而并查集就比较麻烦，细节较多，仔细想想写对就好。

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

struct node
{
int b, w, f[808];
void init (int n)
{
b = w = 0;
for (int i = 0; i <= n; i++)
f[i] = i;
}
int find (int x)
{
if (f[x] == x) return x;
return f[x] = find(f[x]);
}
void merge (int a, int b)
{
f[find(a)] = find(b);
}
};

int n, m;
bool map[205][205];

struct SegmentTree
{
node T[808];
#define LChild p<<1,l,mid
#define RChild p<<1|1,mid+1,r

void maintain (int p, int l, int r)
{
int lc = p<<1, rc = lc|1, mid = (l+r)/2;
T[p].init(n<<2);
T[p].b = T[lc].b + T[rc].b;
T[p].w = T[lc].w + T[rc].w;
for (int i = 1; i <= n; i++)
T[p].f[i] = T[lc].f[i],
T[p].f[i+n] = T[lc].f[i+n],
T[p].f[i+n+n] = T[rc].f[i]+n*2,
T[p].f[i+n+n+n] = T[rc].f[i+n]+n*2;
for (int i = 1; i <= n; i++)
if (map[mid][i]==map[mid+1][i])
{
int f1 = T[p].find(i+n);
int f2 = T[p].find(i+n+n);
if (f1 != f2)
{
T[p].w -= (map[mid][i]==0),
T[p].b -= (map[mid][i]==1);
T[p].merge(f1,f2);
}
}
for (int i = 1; i <= n; i++)
{
int fi = T[p].find(i);
if (fi > n) T[p].f[fi] = T[p].f[i] = i;
}
for (int i = n*3+1; i <= n*4; i++)
{
int fi = T[p].find(i);
if (fi <= n) T[p].f[i-n*2] = T[p].f[i];
else T[p].f[fi] = T[p].f[i-n*2] = i-n*2;
}
}

void make (int r, int p)
{
T[p].init(n<<2);
for (int i = 1; i <= n; i++)
T[p].merge(i,i+n);
for (int i = 2; i <= n; i++)
if (map[r][i]==map[r][i-1])
{
T[p].merge(i,i-1);
T[p].merge(i+n,i+n-1);
}
for (int i = 1; i <= n; i++)
if (T[p].f[i] == i)
T[p].w += map[r][i]==0, T[p].b += map[r][i]==1;
for (int i = 1; i <= n; i++)
if (T[p].f[i+n] == i+n)
T[p].w += map[r][i]==0, T[p].b += map[r][i]==1;
}

void build (int p, int l, int r)
{
if (l == r) make(r,p);
else
{
int mid = (l+r)/2;
build(LChild);
build(RChild);
maintain(p,l,r);
}
}

void change (int p, int l, int r, int pos)
{
if (l == r) {make(r,p);return;}
int mid = (l+r)/2;
if (pos <= mid) change(LChild,pos);
else change(RChild,pos);
maintain(p,l,r);
}

void change (int x, int y)
{
map[x][y] = !map[x][y];
change (1, 1, n, x);
}

void query ()
{
printf ("%d %d\n", T[1].b, T[1].w);
}

}Solve;

int main ()
{
scanf ("%d", &n);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
scanf ("%d", map[i]+j);

Solve.build(1,1,n);
scanf ("%d", &m);
for (int i = 1; i <= m; i++)
{
int x, y; scanf ("%d %d", &x, &y);
Solve.change(x,y);
Solve.query();
}
return 0;
}