codeforces 707D Persistent Bookcase

题意：一个n*m的矩阵执行四种操作：

1，i，j位置置1；

2，i，j位置置1；

3，第i行，0变1，1变0；

4，返回第i次操作的状态。

思路：我还真以为是可持久化呢！我就说CF咋出这么难的数据结构，没想到就一个简单的dfs。

代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<vector>
#include<iostream>
#include<complex>
#include<string>
#include<set>
#include<map>
#include<algorithm>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define nn 100200
#define mm 400200
#define ll long long
#define ULL unsiged long long
#define pb push_back
#define mod 1000000007
#define inf 0x3fffffff
#define eps 0.00000001

int n, m, q;
int op[nn], x[nn], y[nn];
bool a[1010][1010];
int nans = 0;
int ans[nn];
int head[nn], sz=0;
struct Edge {
int to, next, len;
Edge() {}
Edge(int to, int next, int len) :to(to), next(next), len(len) {}
}edge[nn * 2];
void add_edge(int x, int y, int z)
{
edge[sz] = Edge(y, head[x], z);
head[x] = sz++;
}

void dfs(int u)
{
int fuck = 0;
if (op[u] == 1 && a[x[u]][y[u]] == 0) { fuck = 1; a[x[u]][y[u]] = 1; nans++; }
if (op[u] == 2 && a[x[u]][y[u]] == 1) { fuck = 1; a[x[u]][y[u]] = 0; nans--; }
if (op[u] == 3)
{
fuck = 1;
int k = x[u];
for (int i = 1; i <= m; i++)
{
if (a[k][i]) nans--, a[k][i] = 0;
else nans++, a[k][i] = 1;
}
}
ans[u] = nans;
for (int i = head[u]; i!=-1; i = edge[i].next) dfs(edge[i].to);
if (!fuck) return;
if (op[u] == 1) a[x[u]][y[u]] = 0, nans--;
if (op[u] == 2) a[x[u]][y[u]] = 1, nans++;
if (op[u] == 3)
{
int k = x[u];
for (int i = 1; i <= m; i++)
{
if (a[k][i]) nans--, a[k][i] = 0;
else nans++, a[k][i] = 1;
}
}
}
int main()
{
memset(head, -1, sizeof(head));
scanf("%d%d%d", &n, &m, &q);
for (int i = 1; i <= q; i++)
{
scanf("%d%d", &op[i], &x[i]);
if (op[i] == 1 || op[i] == 2) scanf("%d", &y[i]);
if (op[i] == 4) add_edge(x[i], i,1);
else add_edge(i - 1, i,1);
}
for (int i = head[0]; i != -1; i = edge[i].next) dfs(edge[i].to);
for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
return 0;
}