NOIP模拟测试19

数据好水，暴力干翻正解（（（

Problem A: count

一个能被分成大小为k的联通块，必定满足k | n。预处理每个子树大小，枚举n的因子。

我本来以为是树形DP，但颓不出来柿子Orz，把暴力交了结果A了。。

这个做法很容易被卡到T飞：一个长度有巨多因子的链就可以做到。但数据真的很水。。。。

 

#include <bits/stdc++.h>

const int N = 20000005;
int n, ans, ecnt;
int siz[N], head[N];
struct Edge {int to, nxt;} e[N << 1];

inline int read() {
    int a = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) a = a * 10 + c - '0', c = getchar();
    return a;
}

inline void addEdge(int f, int to) {
    e[++ecnt] = {to, head[f]}, head[f] = ecnt;
    e[++ecnt] = {f, head[to]}, head[to] = ecnt;
}

void dfs(int x, int fa) {
    siz[x] = 1;
    for (int i = head[x], y = e[i].to; i; i = e[i].nxt, y = e[i].to) {
        if (y != fa) {
            dfs(y, x);
            siz[x] += siz[y];
        }
    }
}

signed main() {
    n = read();
    for (int i = 1; i < n; i++) {
        int x = read(), y = read();
        addEdge(x, y);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {
            int tmp = 0;
            for (int j = 1; j <= n; j++) {
                if (siz[j] % i == 0) tmp++;
            }
            if (tmp == n / i) ans++;
        }
    }
    return !printf("%d\n", ans);
}

 

Problem B: dinner

破环为链，找出这条链上长度为n的一段

裸暴力是N^3的，考虑优化，我写了二分套二分，加一点比较tricky的小剪枝。

正确性其实我是不敢确定的，因为我也没证。但WA了总比T了好（逃

 

#include <bits/stdc++.h>

int n, m, mx, sum;
int t[100005], pre[100005];

inline int read() {
    int a = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) a = a * 10 + c - '0', c = getchar();
    return a;
}

bool check(int x) {
    for (int i = n + 1, tot = 0; i >= 1 && tot + t[i] <= x; tot += t[i--]){
        int p = i - 1, div = 0;
        while (p < i + n - 1) {
            int l = p + 1, r = i + n - 1, mid, ans;
            while (l <= r) {
                mid = (l + r) >> 1;
                if (pre[mid] - pre[p] <= x)
                    ans = mid, l = mid + 1;
                else r = mid - 1;
            }
            p = ans;
            ++div;
            if (div > m) break;
        }
        if (p == i + n - 1 && div <= m) return true;
    }
    return false;
}

signed main() {
    n = read(), m = read();
    for (int i = 1; i <= n; i++) {
        t[i] = read(), t[i + n] = t[i];
        sum += t[i];
        mx = std::max(mx, t[i]);
    }
    for (int i = 1; i <= 2 * n; i++)
        pre[i] = pre[i - 1] + t[i];
    int l = mx, r = sum, mid, ans = 0;
    while (l <= r) {
        mid = (l + r) >> 1;
        if (check(mid)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    return !printf("%d\n", ans);
}

 

Problem C: chess

考场5min敲了个DFS骗10分，性价比真高啊（

正解：建图跑最短路计数。

但题目中有个条件：两条路径只有经过的空格有不一样才能被认为是不同的。

于是就出现零环的问题，它没用。

怎么解决？我们扫一遍图，对于权值为1的点dfs一遍去连边，这样可以避免零环，连出有效的边。

这题数据锅了，这个做法当时只有70pt。。。换数据后就能A了XD

 

#include <bits/stdc++.h>
#define id(x, y) ((x - 1) * m + y)

const int N = 300003;
const int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int n, m, st, ed;
long long dis[N], cnt[N];
int matrix[505][505], mp[2505][2505];
bool vis[N];
std::vector<int> v;

inline int read() {
    int a = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) a = a * 10 + c - '0', c = getchar();
    return a;
}

void dfs(int x, int y) {
    vis[id(x, y)] = 1;
    for (int i = 0; i <= 7; i++) {
        int xx = x + dx[i], yy = y + dy[i];
        if (xx < 1 || yy < 1 || xx > n || yy > n || vis[id(xx, yy)]) continue;
        if (matrix[xx][yy] != 1 && matrix[xx][yy] != 2) {
            v.push_back(id(xx, yy));
            vis[id(xx, yy)] = 1;
        } else if (matrix[xx][yy] == 1) dfs(xx, yy);
    }
}

void bfs() {
    memset(dis, 0x3f, sizeof(dis));
    dis[st] = 0, cnt[st] = 1;
    std::queue<int> q;
    q.push(st);
    while (!q.empty()) {
        int x = q.front();
        q.pop();
        for (int i = 1; i <= n * m; i++) {
            if (mp[x][i]) {
                if (dis[i] > dis[x] + 1) {
                    if (dis[i] == 0x3f3f3f3f3f3f3f3f) q.push(i);
                    dis[i] = dis[x] + 1;
                    cnt[i] = cnt[x];
                } else if (dis[i] == dis[x] + 1) cnt[i] += cnt[x];
            }
        }
    }
}

signed main() {
    n = read(), m = read();
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            matrix[i][j] = read();
            if (matrix[i][j] == 3) st = id(i, j);
            if (matrix[i][j] == 4) ed = id(i, j);
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (matrix[i][j] == 1 && !vis[id(i, j)]) {
                dfs(i, j);
                for (int x = 0; x < v.size(); x++) {
                    for (int y = x + 1; y < v.size(); y++) {
                        mp[v[x]][v[y]] = mp[v[y]][v[x]] = 1;
                    }
                }
                for (auto x : v) vis[x] = 0;
                v.clear();
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (matrix[i][j] == 1 || matrix[i][j] == 2) continue;
            for (int k = 0; k <= 7; k++) {
                int x = i + dx[k], y = j + dy[k];
                if (x < 1 || y < 1 || x > n || y > m) continue;
                if (matrix[x][y] == 2 || matrix[x][y] == 1) continue;
                mp[id(i, j)][id(x, y)] = mp[id(x, y)][id(i, j)] = 1;
            }
        }
    }
    bfs();
    if (dis[ed] == 0x3f3f3f3f3f3f3f3f) {
        puts("-1");
    } else {
        printf("%lld\n%lld\n", dis[ed] - 1, cnt[ed]);
    }
    return 0;
}