HNOI2013解题报告

HNOI2013解题报告

Author: Pengyihao

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Day1 T1 比赛

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思路

这是一道搜索的题目。

一个重要的优化就是，因为球队的分数排序后是不影响后面的答案的，所以判重的时候可以很方便。

然后还有就是 2810 是不会爆 long long 的……

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代码

#include <bits/stdc++.h>typedef long long LL;#define FOR(i, a, b) for (int i = (a), i##_END_ = (b); i <= i##_END_; i++)#define DNF(i, a, b) for (int i = (a), i##_END_ = (b); i >= i##_END_; i--)template <typename Tp> void in(Tp &x) {    char ch = getchar(), f = 1; x = 0;    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();    if (ch == '-') f = -1, ch = getchar();    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();    x *= f;}template <typename Tp> Tp chkmax(Tp &x, Tp y) {return x > y ? x : x=y;}template <typename Tp> Tp chkmin(Tp &x, Tp y) {return x < y ? x : x=y;}template <typename Tp> Tp Max(Tp x, Tp y) {return x > y ? x : y;}template <typename Tp> Tp Min(Tp x, Tp y) {return x < y ? x : y;}const int MAXN = 20, MOD = 1000000007;LL a[MAXN];std::map<LL, LL>mp;LL hash(int now){    LL res = now, tmp[MAXN];    FOR(i, 1, now) tmp[i] = a[i];    std::sort(tmp + 1, tmp + now + 1, std::less<int>());    FOR(i, 1, now) res += res * 28 + tmp[i];    return res;}LL DFS(int now, int n){    if (a[n] > 3 * (n - now)) return -1;    LL res = 0;    if (now == n) {        if (n == 1) return 1;        else {            LL h = hash(n - 1);            if (mp[h]) return mp[h];            return mp[h] = DFS(1, n - 1);        }    }    if (a[n] >= 3) {        LL tmp = 0;        a[n] -= 3, tmp = DFS(now + 1, n);        if (tmp != -1) (res += tmp) %= MOD;        a[n] += 3;    }    if (a[n] && a[now]) {        LL tmp = 0;        a[n]--, a[now]--, tmp = DFS(now + 1, n);        if (tmp != -1) (res += tmp) %= MOD;        a[n]++, a[now]++;    }    if (a[now] >= 3) {        LL tmp = 0;        a[now] -= 3, tmp = DFS(now + 1, n);        if (tmp != -1) (res += tmp) %= MOD;        a[now] += 3;    }    return res ? res : -1;}int n;int main(){    freopen("match.in", "r", stdin);    freopen("match.out", "w", stdout);    in(n);    FOR(i, 1, n) in(a[i]);    std::sort(a + 1, a + n + 1, std::less<int>());    printf("%lld\n", DFS(1, n));    return 0;}

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Day1 T2 消毒

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思路

可以发现最优的话一定要是，选择的区域中的一维为 1，另外两维为最大值。

因为 abc≤5000，所以它们中的最小值是小于 20 的。

把这一维看作高，然后枚举每一层选不选。

然后剩下的就转化成了二维问题。

这就是个典型的二分图最小点覆盖，网络流跑就可以了。

时间复杂度为 O(松) 。注意要进行常数优化。

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代码

#include <bits/stdc++.h>typedef long long LL;#define debug(...) fprintf(stderr, __VA_ARGS__)#define FOR(i, a, b) for (int i = (a), i##_END_ = (b); i <= i##_END_; i++)#define DNF(i, a, b) for (int i = (a), i##_END_ = (b); i >= i##_END_; i--)template <typename Tp> void in(Tp &x) {    char ch = getchar(), f = 1; x = 0;    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();    if (ch == '-') f = -1, ch = getchar();    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();    x *= f;}template <typename Tp> Tp chkmax(Tp &x, Tp y) {return x > y ? x : x=y;}template <typename Tp> Tp chkmin(Tp &x, Tp y) {return x < y ? x : x=y;}template <typename Tp> Tp Max(Tp x, Tp y) {return x > y ? x : y;}template <typename Tp> Tp Min(Tp x, Tp y) {return x < y ? x : y;}using std::vector;const int MAXN = 5002;bool chose[20];int a[4], ans;bool map[MAXN][MAXN];std::vector<std::vector<int> >g[MAXN];void init(){    ans = 0x3f3f3f3f;    FOR(i, 1, 3) in(a[i]);    FOR(i, 0, a[1] + 1) {        g[i].resize(a[2] + 2);        FOR(j, 0, a[2] + 1) g[i][j].resize(a[3] + 2);    }    FOR(i, 1, a[1]) {        FOR(j, 1, a[2]) FOR(k, 1, a[3]) in(g[i][j][k]);    }}const int ss = 0, tt = 5008, INF = 0x3f3f3f3f;using std::queue;queue<int>q;bool link[MAXN];int dis[5010], cur[5010];int cnt, head[5010], data[20010], flow[20010], nxt[20010];void add(int x, int y, int z){    nxt[cnt] = head[x]; data[cnt] = y; flow[cnt] = z; head[x] = cnt++;    nxt[cnt] = head[y]; data[cnt] = x; flow[cnt] = 0; head[y] = cnt++;}int times, maxx = 1, alledge;struct Edge {    int x, y;    Edge(int a=0, int b=0): x(a), y(b) {}};std::vector<Edge>edge[20];bool bfs(){    dis[ss] = times; q.push(ss);    while (!q.empty()) {        int now = q.front(); q.pop();        for (int i = head[now]; i != -1; i = nxt[i]) {            if (dis[data[i]] < times && flow[i]) {                dis[data[i]] = dis[now] + 1;                q.push(data[i]);                chkmax(maxx, dis[now] + 1);            }        }    }    return dis[tt] >= times;}int dfs(int now, int fl){    if (now == tt) return fl;    int flo;    for (int i = head[now]; i != -1; i = nxt[i]) {        if (flow[i] && dis[data[i]] == dis[now] + 1) {            if (flo = dfs(data[i], Min(fl, flow[i]))) {                flow[i] -= flo;                flow[i ^ 1] += flo;                return flo;            }        }    }    return 0;}void check(int minx){    int tmptot = 0;    FOR(i, 1, minx) if (chose[i]) tmptot++;    if (tmptot >= ans) return;    cnt = 0;    int tot = tmptot;// ***********************************************// make_Edge    int Addition;    if (minx == a[1]) Addition = a[2];    else if (minx == a[2]) Addition = a[1];    else Addition = a[1];    FOR(k, 1, minx) if (!chose[k]) {        FOR(i, 0, edge[k].size() - 1) {            int fr = edge[k][i].x, to = edge[k][i].y + Addition;            if (!map[fr][to - Addition]) {                add(fr, to, 1);                if (!link[fr]) {link[fr] = true; add(ss, fr, 1);}                if (!link[to]) {link[to] = true; add(to, tt, 1);}            }        }    }// ***********************************************    int fl = 0;    times = maxx + 1;    while (bfs()) {        //memcpy(cur, head, sizeof head);        int tmp;        while (tmp = dfs(ss, INF)) fl += tmp;        times = maxx + 1;    }    chkmin(ans, fl + tot);    FOR(k, 1, minx) if (!chose[k]) FOR(i, 0, edge[k].size() - 1) {        map[edge[k][i].x][edge[k][i].y] = false;        link[edge[k][i].x] = link[edge[k][i].y + Addition] = false;        head[edge[k][i].x] = head[edge[k][i].y + Addition] = -1;    }    head[ss] = head[tt] = -1;}void search(int now, int minx){    chose[now] = false;    if (now == minx) check(minx);    else search(now + 1, minx);    chose[now] = true;    if (now == minx) check(minx);    else search(now + 1, minx);    chose[now] = false;}void work(){    memset(head, -1, sizeof head);    int minx = 5000;    FOR(i, 1, 3) chkmin(minx, a[i]);    alledge = 0;    FOR(i, 1, minx) {        edge[i].clear();    }    if (minx == a[1]) {        FOR(k, 1, a[1]) {            FOR(i, 1, a[2])    FOR(j, 1, a[3])                if (g[k][i][j]) edge[k].push_back(Edge(i, j));        }    }    else if (minx == a[2]) {            FOR(i, 1, a[1]) FOR(k, 1, a[2]) { FOR(j, 1, a[3])                if (g[i][k][j]) edge[k].push_back(Edge(i, j));        }    }    else {            FOR(i, 1, a[1]) FOR(j, 1, a[2]) FOR(k, 1, a[3]) {                if (g[i][j][k]) edge[k].push_back(Edge(i, j));            }    }    search(1, minx);    printf("%d\n", ans);    debug("%d\n", ans);}int main(){    freopen("clear.in", "r", stdin);    freopen("clear.out", "w", stdout);    int tcase; in(tcase);    while (tcase--) {        init();        work();    }    return 0;}

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Day1 T3 旅行

这是一道很难的题目。我看了题解也暂时无法解决此题，所以决定跳过它。

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Day2 T1 数列

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思路

考虑差分数列。

差分数列的可能数量为 mk−1。

于是答案就为 ∑n−∑k−1i=1ai

其中 ai 表示差分数列的第 i 项，最前面的 ∑ 表示枚举的差分数列。

n 可以提出来，而 ai 的贡献是独立的。

所以最后答案为

mk−1×n−m(m+1)/2×(k−1)×mk−2

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代码

#include <bits/stdc++.h>typedef long long LL;#define FOR(i, a, b) for (int i = (a), i##_END_ = (b); i <= i##_END_; i++)#define DNF(i, a, b) for (int i = (a), i##_END_ = (b); i >= i##_END_; i--)template <typename Tp> void in(Tp &x) {    char ch = getchar(), f = 1; x = 0;    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();    if (ch == '-') f = -1, ch = getchar();    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();    x *= f;}template <typename Tp> Tp chkmax(Tp &x, Tp y) {return x > y ? x : x=y;}template <typename Tp> Tp chkmin(Tp &x, Tp y) {return x < y ? x : x=y;}template <typename Tp> Tp Max(Tp x, Tp y) {return x > y ? x : y;}template <typename Tp> Tp Min(Tp x, Tp y) {return x < y ? x : y;}LL ans = 0;LL n, k, m, p;LL power(LL x, LL y, LL p){    x %= p;    LL ret = 1;    while (y) {        if (y & 1) ret = 1ll * ret * x % p;        x = 1ll * x * x % p;        y >>= 1;    }    return ret;}int main(){    freopen("seq.in", "r", stdin);    freopen("seq.out", "w", stdout);    in(n); in(k); in(m); in(p);    n %= p;    ans = 1ll * power(m, k - 1, p) * n % p;    ans = ((ans - 1ll * (1ll * m * (m + 1) / 2) % p *           (k - 1) % p * power(m, k - 2, p) % p) % p + p) % p;    printf("%lld\n", ans);    return 0;}

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Day2 T2 游走

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思路

我们可以贪心来做——求出每条边走过的期望次数，然后从大到小分配从 1 到 m 的编号。

边的期望次数的求法是一个裸的高斯消元，这里就不再赘述。

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代码

#include <bits/stdc++.h>const int MAXN = 510;bool f[MAXN][MAXN];int n, m, du[MAXN], all;long double matrix[MAXN][MAXN], times[MAXN], timess[MAXN * MAXN];template <typename Tp> void in(Tp &x) {    char ch = getchar(); x = 0;    while (ch < '0' || ch > '9') ch = getchar();    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();}void gauss() {    for (int j = 1; j <= n; j++) {        long double maxu = -1;        int maxv = 0;        for (int i = j; i <= n + 1; i++)            if (maxu < fabs(matrix[i][j])) {                maxu = fabs(matrix[i][j]);                maxv = i;            }        for (int i = 1; i <= n + 1; i++) {            long double t = matrix[j][i];            matrix[j][i] = matrix[maxv][i];            matrix[maxv][i] = t;        }        long double eps = matrix[j][j];        if (fabs(eps) < 1e-10) continue;        for (int i = 1; i <= n + 1; i++)            matrix[j][i] /= eps;        for (int i = 1; i <= n + 1; i++)            if (i != j) {                long double epss = matrix[i][j];                if (fabs(epss) < 1e-10) continue;                for (int k = 1; k <= n + 1; k++)                    matrix[i][k] -= matrix[j][k] * epss;            }    }    for (int i = 1; i <= n; i++)        times[i] = matrix[i][n + 1] / matrix[i][i];    for (int i = 1; i <= n; i++)        for (int j = i + 1; j <= n; j++)            if (f[i][j]) {                if (j == n) timess[++all] = times[i] / du[i];                else timess[++all] = times[i] / du[i] + times[j] / du[j];            }    std::sort(timess + 1, timess + all + 1);    long double ans = 0;    for (int i = 1; i <= all; i++) {        ans += timess[i] * (all - i + 1);    }    printf("%.3Lf\n", ans);}int main() {    freopen("walk.in", "r", stdin);    freopen("walk.out", "w", stdout);    in(n); in(m);    for (int i = 1; i <= m; i++) {        int u, v;        in(u); in(v);        f[u][v] = f[v][u] = true;        if (u != n) du[u]++;        if (v != n) du[v]++;    }    for (int i = 1; i <= n; i++) {        for (int j = 1; j < n; j++)            if (f[i][j])                matrix[i][j] = (long double)(1) / du[j];        matrix[i][i] = -1;        if (i == 1) matrix[i][n + 1] = -1;    }    matrix[n + 1][n] = 1;    matrix[n + 1][n + 1] = 1;    gauss();    return 0;}

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Day2 T3 切糕

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思路

对于每一个位置，从所有层向上一层连边，容量为点权。

对于每一个位置，从第 i 层向相邻位置的 i−d 层连边，容量为 +∞。

对于每一个位置，从源点向第一层连边，容量为 +∞。

对于每一个位置，从最高层向汇点连边，容量为 +∞。

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代码

#include <bits/stdc++.h>#define PLA(i, j, k) ((i) * 2500 + (j) * 50 + (k))typedef long long LL;#define FOR(i, a, b) for (int i = (a), i##_END_ = (b); i <= i##_END_; i++)#define DNF(i, a, b) for (int i = (a), i##_END_ = (b); i >= i##_END_; i--)template <typename Tp> void in(Tp &x) {    char ch = getchar(), f = 1; x = 0;    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();    if (ch == '-') f = -1, ch = getchar();    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();    x *= f;}template <typename Tp> Tp chkmax(Tp &x, Tp y) {return x > y ? x : x=y;}template <typename Tp> Tp chkmin(Tp &x, Tp y) {return x < y ? x : x=y;}template <typename Tp> Tp Max(Tp x, Tp y) {return x > y ? x : y;}template <typename Tp> Tp Min(Tp x, Tp y) {return x < y ? x : y;}int p, qs, r, d;int height[50][50][50];int cnt, dis[200010], head[200010], data[200010 << 1], flow[200010 << 1], cur[200010], nxt[200010 << 1];using std::queue; queue<int>q;const int ss = 0, tt = 200009, INF = 0x3f3f3f3f;bool bfs(){    memset(dis, -1, sizeof dis);    dis[ss] = 0; q.push(ss);    while (!q.empty()) {        int now = q.front(); q.pop();        for (int i = head[now]; i != -1; i = nxt[i]) {            if (dis[data[i]] == -1 && flow[i]) {                dis[data[i]] = dis[now] + 1;                q.push(data[i]);            }        }    }    return dis[tt] != -1;}int dfs(int now, int fl){    if (now == tt) return fl;    int flo;    for (int &i = cur[now]; i != -1; i = nxt[i]) {        if (flow[i] && dis[data[i]] == dis[now] + 1) {            if (flo = dfs(data[i], Min(fl, flow[i]))) {                flow[i] -= flo;                flow[i ^ 1] += flo;                return flo;            }        }    }    return 0;}void add(int x, int y, int z){    nxt[cnt] = head[x]; data[cnt] = y; flow[cnt] = z; head[x] = cnt++;    nxt[cnt] = head[y]; data[cnt] = x; flow[cnt] = 0; head[y] = cnt++;}int main(){    freopen("cake.in", "r", stdin);    freopen("cake.out", "w", stdout);    memset(head, -1, sizeof head);    in(p); in(qs); in(r); in(d);    FOR(i, 1, r) {        FOR(j, 1, p) FOR(k, 1, qs) in(height[i][j][k]);    }    FOR(i, 1, p) FOR(j, 1, qs) {        add(ss, PLA(i, j, 1), INF);        FOR(k, 1, r) {            add(PLA(i, j, k), PLA(i, j, k + 1), height[k][i][j]);        }        add(PLA(i, j, r + 1), tt, INF);    }    FOR(i, 1, p) FOR(j, 1, qs) {        FOR(k, d + 1, r + 1) {            if (i != p) add(PLA(i, j, k), PLA(i + 1, j, k - d), INF);            if (i != 1) add(PLA(i, j, k), PLA(i - 1, j, k - d), INF);            if (j != qs) add(PLA(i, j, k), PLA(i, j + 1, k - d), INF);            if (j != 1) add(PLA(i, j, k), PLA(i, j - 1, k - d), INF);        }    }    int fl = 0;    while (bfs()) {        memcpy(cur, head, sizeof head);        int tmp; while (tmp = dfs(ss, INF)) fl += tmp;    }    printf("%d\n", fl);    return 0;}