CF368 D

re了20多发 还是我在测试数据上操作最后了10多发才发现的

其实只需要多加一句就好了

真的愚蠢啊，要不都能进前100了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

int sh[1005][1005];
int flag[1005];
int sum[1005];
int all;
int vis[MAXN];
vector<int> mp[MAXN];
int n,m,q;

struct Node{
	int ty; int x,y; int ans;
}as[MAXN];

int cc = 0;
const int Ed = 26;
void dfs(int po,int fa) {
	
	int x1,x2; int y1,y2,y3,y4; // sh[x1][x2], all, sum[x1], flag[x1]	
	for(int i = 0; i < (int)mp[po].size(); ++i) {
		int tt = mp[po][i];
		int t1 = as[tt].ty; int t2 = as[tt].x; int t3 = as[tt].y;
			
		if(t1 == 4) t2 = 0; // 这地方不写会越界 谨记啊
		x1 = t2; x2 = t3; y1 = sh[x1][x2]; y2 = all; y3 = sum[x1]; y4 = flag[x1];
	
		if(t1 == 1) {
			if( (flag[t2] ^ sh[t2][t3]) == 0) {
				all ++; sh[t2][t3] ^= 1; sum[t2] ++;
			}
		}else if(t1 == 2) {
			if( (flag[t2] ^ sh[t2][t3]) == 1){
				all--; sh[t2][t3] ^= 1; sum[t2] --;
			}
		}else if(t1 == 3) {
			flag[t2] ^= 1;
			all += m-sum[t2]*2;
		   	sum[t2] = m-sum[t2];
		}
		
		as[tt].ans = all;
		dfs(tt,po);
		
		sh[x1][x2] = y1; all = y2; sum[x1] = y3; flag[x1] = y4;
	}
}
int main(){
	while(~scanf("%d %d %d",&n,&m,&q)){
		memset(flag,0,sizeof(flag));
		memset(sh,0,sizeof(sh));
		memset(sum,0,sizeof(sum));
		all = 0;
		cc = 0;

		for(int i = 0; i <= q; ++i) mp[i].clear();

		for(int i = 1; i <= q; ++i) {
			as[i].y = 0;
			scanf("%d",&as[i].ty);
			if(as[i].ty < 3) scanf("%d %d",&as[i].x,&as[i].y);
			else scanf("%d",&as[i].x);

			if(as[i].ty == 4) mp[as[i].x].push_back(i);
			else mp[i-1].push_back(i);
		}

		dfs(0,-1);
		for(int i = 1; i <= q; ++i) printf("%d
",as[i].ans);
	}
	return 0;
}