wenbao与cf书架放书

题目：

http://codeforces.com/contest/707/problem/D

将书架与层数分别与时间建立关系

大神代码

#include <bits/stdc++.h>
using namespace std;
#define N 1010
#define M 100100

int n, m, q, type, a, b;
int last[M][N], f[M];
bool on[M][N];

int main() {
    scanf("%d %d %d", &n, &m, &q);
    for (int t = 1; t <= q; t ++) {
        for (int i = 1; i <= n; i ++) last[t][i] = last[t-1][i];
        scanf("%d", &type);
        if (type == 1 || type == 2) {
            scanf("%d %d", &a, &b);
f[t] = f[t-1];
            for (int j = 1; j <= m; j ++) on[t][j] = on[last[t][a]][j];
            last[t][a] = t;
            f[t] -= on[t][b];
            on[t][b] = (type == 1);
            f[t] += on[t][b];
        }
        else if (type == 3) {
            scanf("%d", &a);
            f[t] = f[t-1];
            for (int j = 1; j <= m; j ++) {
                f[t] -= on[last[t][a]][j];
                on[t][j] = !on[last[t][a]][j];
                f[t] += on[t][j];
            }
            last[t][a] = t;
        }
        else {
            scanf("%d", &a);
            f[t] = f[a];
            for (int i = 1; i <= n; i ++) last[t][i] = last[a][i];
        }
        printf("%d
", f[t]);
    }
    return 0;
}

弱鸡代码

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3+10;
const int maxn2 = 1e5+10;
int T[maxn][maxn]; // T[i][j] 第i次处理的第j个书架
bool Q[maxn][maxn]; // Q[i][j] 第i次处理的第j层
int num[maxn2]; //num[i] 第i次处理的书的个数
int main()
{
    int n,m,q,a,b,t;
    cin>>n>>m>>q;
    for(int i = 1; i <= q; i ++){
        for(int j = 1; j <= n; j ++) T[i][j] = T[i-1][j];
        cin>>t;
        if(t == 1 || t == 2){
            num[i] = num[i-1];
            cin>>a>>b;
            for(int k = 1; k <= m; k ++) Q[i][k] = Q[T[i][a]][k];
            T[i][a] = i;
            num[i] -= Q[i][b];
            Q[i][b] = (t == 1);
            num[i] += Q[i][b];
        }
        else if(t == 3){
            num[i] = num[i-1];
            cin>>a;
            for(int k = 1; k <= m; k ++){
                num[i] -= Q[T[i][a]][k];
                Q[i][k] =! Q[T[i][a]][k];
                num[i] += Q[i][k];
            }
            T[i][a] = i;
        }
        else{
            cin>>a;
            num[i] = num[a];
            for(int j = 1; j <= n; j ++) T[i][j] = T[a][j];
        }
        cout<<num[i]<<endl;
    }
}
/*  13
7 2 18
2 5 2
1 3 1
2 7 1
3 4
4 0
1 6 1
4 0
4 4
1 6 2
4 7
3 5
2 2 2
2 6 1
4 7
1 7 2
2 4 2
3 4
1 1 1
Output
0
1
1
3
0
1
0
3
4
0
2
2
2
0
1
1
3
4
*/
/* 15
19 7 17
1 7 3
2 6 5
2 16 3
2 11 3
3 2
1 18 1
1 14 1
3 6
4 4
2 13 3
2 16 6
3 18
1 4 7
4 2
1 9 4
4 8
4 10
Output
1
1
1
1
8
9
10
17
1
1
1
8
9
1
2
17
1
*/

只有不断学习才能进步！