题目链接：http://codeforces.com/contest/707/problem/D
题意：维护一个图，一共四个操作。
1.第i行第j列变1
2.第i行第j列变0
3.第i行颠倒，1变0，0变1
4.将整张图变为k次操作时的状态如果k是0的话，则清空图
乍一看就知道可以用bitset来保存图，但是更新图的话复杂度太高，纯模拟一定会TLE。所以可以离线做，存下所有状态。并且根据操作建树，假如j操作是i次操作的下一步，则将j操作作为i的儿子，1~3操作都是i-1作为i的父亲，只有4操作特殊，设i是k的儿子。这里树根设0，0存一个空的图就解决操作4的操作数是0的问题了。之后回溯每次查询，查询完一条后回溯到前一状态走另外的儿子就行了。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onecnt(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef pair<LL, LL> pll; 65 typedef map<string, int> msi; 66 typedef vector<int> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 typedef struct Node { 72 int k, x, y; 73 }Node; 74 const int maxn = 1010; 75 const int maxm = 100100; 76 int ret; 77 int q, n, m; 78 int cmd, a, b, c; 79 bitset<maxn> st[maxn]; 80 vi mp[maxm]; 81 Node qq[maxm]; 82 int ans[maxm]; 83 84 void go(int x, int& ok) { 85 int k = qq[x].k; 86 int i = qq[x].x; 87 int j = qq[x].y; 88 if(k == 1) { 89 if(!st[i][j]) { 90 st[i][j] = 1; 91 ret++; 92 ok = 1; 93 } 94 else ok = 0; 95 } 96 if(k == 2) { 97 if(st[i][j]) { 98 st[i][j] = 0; 99 ret--; 100 ok = 1; 101 } 102 else ok = 0; 103 } 104 if(k == 3) { 105 int cur = st[i].count(); 106 if(st[i][maxn-1] == 1) cur -= (maxn - m); 107 ret = ret - cur + (m - cur); 108 st[i] = ~st[i]; 109 } 110 } 111 112 void re(int x, int& ok) { 113 int k = qq[x].k; 114 int i = qq[x].x; 115 int j = qq[x].y; 116 if(k == 1) { 117 if(ok) { 118 st[i][j] = 0; 119 ret--; 120 } 121 } 122 if(k == 2) { 123 if(ok) { 124 st[i][j] = 1; 125 ret++; 126 } 127 } 128 if(k == 3) { 129 int cur = st[i].count(); 130 if(st[i][maxn-1] == 1) cur -= (maxn - m); 131 ret = ret - cur + (m - cur); 132 st[i] = ~st[i]; 133 } 134 } 135 136 void dfs(int x) { 137 int ok; 138 if(x) { 139 go(x, ok); 140 ans[x] = ret; 141 } 142 Rep(i, mp[x].size()) dfs(mp[x][i]); 143 re(x, ok); 144 } 145 146 int main() { 147 // FRead(); 148 while(~scanf("%d%d%d",&n,&m,&q)) { 149 For(i, 1, q+1) { 150 Rint(qq[i].k); 151 if(qq[i].k <= 2) { 152 Rint(qq[i].x); 153 Rint(qq[i].y); 154 } 155 else Rint(qq[i].x); 156 } 157 For(i, 2, q+1) { 158 if(qq[i].k == 4) mp[qq[i].x].push_back(i); 159 else mp[i-1].push_back(i); 160 } 161 mp[0].push_back(1); 162 dfs(0); 163 For(i, 1, q+1) printf("%d ", ans[i]); 164 } 165 RT 0; 166 }