Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals)

http://codeforces.com/contest/828

C题直接的思路是模拟把整个字符串都覆盖上去，已经覆盖的，就不需要再覆盖就是了。

这个套路出过很多次了，是一个类似并查集的东西，也就是并查集维护线段吧，我直接dfs回溯弄也可以。

设tonext[i]表示第i个位置的下一个空位是什么，

注意中间空的位置填上'a'就好

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 3e6 + 20;
char str[maxn], sub[maxn];
int tonext[maxn], haha;
int mx;
void dfs(int cur, int pos, int lef) {
    mx = max(mx, haha);
    if (lef <= 0) return;
    if (tonext[cur] == cur) {
        str[cur] = sub[pos];
        tonext[cur] = cur + 1;
        haha = cur + 1;
        dfs(cur + 1, pos + 1, lef - 1);
    } else {
        haha = tonext[cur];
        dfs(tonext[cur], pos + tonext[cur] - cur, lef - (tonext[cur] - cur));
        tonext[cur] = haha;
    }
}
void work() {
    for (int i = 1; i <= maxn - 20; ++i) tonext[i] = i;
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> sub + 1;
        int k;
        cin >> k;
        int lensub = strlen(sub + 1);
        for (int j = 1; j <= k; ++j) {
            int pos;
            cin >> pos;
            dfs(pos, 1, lensub);
        }
    }
    for (int i = 1; i < mx; ++i) {
        if (str[i] == '') str[i] = 'a';
    }
    cout << str + 1 << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    IOS;
    work();
    return 0;
}

D题想歪了，要使得最远的距离最短，则分两种情况吧，具体思路都是选出一个点，作为树的重心，然后其他非叶子节点连上去，或则叶子节点连上去。

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

void work() {
    int n, k;
    cin >> n >> k;
    int to = 1;
    if (n - k == 1) {
        printf("2
");
        for (int i = 1; i <= k; ++i) {
            printf("%d %d
", i, n);
        }
    } else if (n - k == 2) {
        printf("3
");
        printf("1 2
");
        printf("1 3
");
        printf("2 4
");
        for (int i = 5; i <= 2 + k; ++i) {
            printf("1 %d
", i);
        }
    } else {
        if (k >= n - k - 1) {
            printf("4
");
            for (int i = k + 2; i <= n; ++i) {
                printf("%d %d
", k + 1, i);
            }
            int to = 0;
            for (int i = k + 2; i <= n; ++i) {
                printf("%d %d
", i, ++to);
            }
            while (to < k) {
                ++to;
                printf("%d %d
", k + 1, to);
            }
        } else {
            int res = (n - k - 1) % k;
            int ans;
            if (res == 0) {
                ans = 2 * (1 + (n - k - 1) / k);
            } else if (res == 1) {
                ans = 2 * (1 + (n - k - 1) / k) + 1;
            } else ans = 2 * (1 + (n - k - 1) / k) + 2;
            printf("%d
", ans);
            for (int i = 1; i <= k; ++i) {
                printf("%d %d
", k + 1, i);
            }
            int one = 1, two = k + 2, pre = k + 2, up = k;
            for (int i = 1; i <= n - k - 1; ++i) {
                printf("%d %d
", one, two);
                one++;
                two++;
                if (one > up) {
                    one = pre;
                    up = two - 1;
                    pre = two;
                }
            }
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

E题我没什么好方法用了220个bit怼过去了。

就是维护跳跃的区间，不知道其他大佬怎么做的。

做这题的时候发现有4个字母和长度最大是10，应该有点用的。

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
int c[5][57][100000 + 2];
int lenstr;
int lowbit(int x) {
    return x & (-x);
}
void upDate(int which, int dis, int pos, int val) {
    while (pos <= lenstr) {
        c[which][dis][pos] += val;
        pos += lowbit(pos);
    }
}
int ask(int which, int dis, int pos) {
    int ans = 0;
    while (pos) {
        ans += c[which][dis][pos];
        pos -= lowbit(pos);
    }
    return ans;
}
char str[111111];
int getID(char ch) {
    if (ch == 'A') return 0;
    else if (ch == 'T') return 1;
    else if (ch == 'G') return 2;
    else return 3;
}
char sub[111];
int num[22][22];
int calc(int pos, int dis) {
    pos %= dis;
    if (pos == 0) pos = dis;
    return num[pos][dis];
}
void work() {
    scanf("%s", str + 1);
    lenstr = strlen(str + 1);
    for (int i = 1; i <= lenstr; ++i) {
        int id = getID(str[i]);
        for (int j = 1; j <= 10; ++j) {
//            if (j > i) break;
            upDate(id, calc(i, j), i, 1);
        }
    }
    int q;
    scanf("%d", &q);
    while (q--) {
        int op;
        scanf("%d", &op);
        if (op == 1) {
            int pos;
            scanf("%d%s", &pos, sub);
            if (str[pos] == sub[0]) continue;
            int id = getID(str[pos]);
            int id2 = getID(sub[0]);
            for (int i = 1; i <= 10; ++i) {
//                if (i > pos) break;
                upDate(id, calc(pos, i), pos, -1);
                upDate(id2, calc(pos, i), pos, 1);
            }
            str[pos] = sub[0];
        } else {
            int L, R;
            scanf("%d%d%s", &L, &R, sub + 1);
            int lensub = strlen(sub + 1);
            int ans = 0;
            if (lensub >= R - L + 1) {
                for (int i = L; i <= R; ++i) {
                    ans += str[i] == sub[i - L + 1];
                }
            } else {
                int newR = R - (L - 1);
                int res = newR % lensub;
                for (int i = 1, to = R - res + 1; i <= res; ++i, ++to) {
                    ans += str[to] == sub[i];
                }
                for (int i = R - res - lensub + 1, to = 1; i <= R - res; ++i, ++to, ++L) {
                    int id = getID(sub[to]);
                    ans += ask(id, calc(i, lensub), i) - ask(id, calc(i, lensub), L) + (str[L] == sub[to]);
//                    printf("%d
", ans);
                }
            }
            printf("%d
", ans);
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int to = 0;
    for (int i = 1; i <= 10; ++i) {
        for (int j = i; j <= 10; ++j) {
            num[i][j] = ++to;
        }
    }
    work();
    return 0;
}