Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3201 [HNOI2009] 梦幻布丁

P3201 [HNOI2009] 梦幻布丁

注意到只有将一个颜色变为另一个颜色的操作,那么颜色的种类一定不会增加,考虑启发式合并。用链表维护每个颜色的位置,每次将颜色数量少的暴力合并到大的。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10, M = 1e6 + 10;
int n, m, ans, c[N], sz[M], p[M];
int idx, hd[M], nxt[N], edg[N];
void merge(int &x, int &y)
{
if (x == y)
return;
if (sz[x] > sz[y])
swap(x, y);
for (int i = hd[x]; ~i; i = nxt[i])
ans -= (c[edg[i] - 1] == y) + (c[edg[i] + 1] == y);
for (int i = hd[x]; ~i; i = nxt[i])
{
c[edg[i]] = y;
if (nxt[i] == -1)
{
nxt[i] = hd[y];
hd[y] = hd[x];
break;
}
}
hd[x] = -1;
sz[y] += sz[x];
sz[x] = 0;
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i < M; i++)
hd[i] = -1;
for (int i = 1; i <= n; i++)
{
scanf("%d", &c[i]);
if (c[i] != c[i - 1])
ans++;
add(c[i], i);
sz[c[i]]++;
}
for (int i = 0; i < M; i++)
p[i] = i;
for (int op; m; m--)
{
scanf("%d", &op);
if (op == 1)
{
int x, y;
scanf("%d%d", &x, &y);
merge(p[x], p[y]);
}
else if (op == 2)
printf("%d\n", ans);
}
return 0;
}