模拟赛小结：2017 China Collegiate Programming Contest Final (CCPC-Final 2017)

比赛链接：传送门

前期大顺风，2：30金区中游。后期开题乏力，掉到银尾。4：59绝杀I，但罚时太高卡在银首。

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Problem A - Dogs and Cages 00：09：45 （+） Solved by Dancepted

算了半天发现就是n-1，被队友喷死，差点气哭。

Problem E - Evil Forest 00：16：54 （+） Solved by xk

xk大喊签到，就过了。

代码：

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

int main() {
    fast;
    int T, kase = 1;
    cin >> T;
    while (T--) {
        int n;
        cin >> n;
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            int x;
            cin >> x;
            ans += x + upperdiv(x, 10);
        }
        cout << "Case #" << (kase++) << ": " << ans << endl;
    }
    return 0;
}

Problem C - Rich Game 00：32：42 （+） Solved by xk

xk又喊了一声签到！

代码：

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

int main() {
    fast;
    int T, kase = 1;
    cin >> T;
    while (T--) {
        int x, y, k;
        cin >> x >> y >> k;
        int money = 0, ans = 0;
        if(x > y) ans = k;
        else
        {
            for(int i = 0; i < k; i++)
            {
                if(money + 9 * x >= 11 * y) {
                    money += 9 * x - 11 * y;
                    ans++;
                }
                else {
                    money += 11 * x;
                }
            }
        }
        cout << "Case #" << (kase++) << ": " << ans << endl;
    }
    return 0;
}

Problem K - Knightmare 00：46：33 （+） Solved by Dancepted

步数较少的时候可以往回走，把没走过的地方填上，步数增长到一定程度时，答案差不多会均匀增长。

我猜测是小数据打表，大数据是个公式之类的。

开完A趁电脑没人就打了个表，发现和我的猜测是一致的，果断交了一发就过了。

代码：

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <stdio.h>
#include <utility>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 1005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef __int128 ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

const int ans[10] = {1, 9, 41, 109, 205, 325};
int main() {
    int T; cin >> T;
    int kase = 1;
    while (T--) {
        ll n; read(n);
        printf("Case #%d: ", kase++);
        if (n <= 5) {
            write(ans[n]);
        }
        else {
            ll res = 14 * n * n - 6 * n + 5;
            write(res);
        }
        putchar('
');
    }
    
    return 0;
}

Problem J - Subway Chasing 01：42：00 （+） Solved by xk & lh

好像是差分约束什么的。图论这种事相信队友就完了。

代码：

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

const int maxn = 2000 + 5;

struct edge
{
    int v, l;
};

ll dis[maxn];
int cnt[maxn];
bool inq[maxn];
vector<edge> g[maxn];

void addedge(int u, int v, int l)
{
    g[u].push_back(edge{v, l});
    // g[v].push_back(edge{u, l});
}

int n, m, x;

bool spfa()
{
    queue<int> q;
    dis[1] = 0;
    q.push(1);
    cnt[1] = inq[1] = 1;
    while(!q.empty())
    {
        int u = q.front();
        inq[u] = 0;
        q.pop();
        for(auto &e : g[u])
        {
            if(dis[e.v] > dis[u] + e.l)
            {
                dis[e.v] = dis[u] + e.l;
                
                if(!inq[e.v]) {
                    cnt[e.v]++;
                    if(cnt[e.v] >= n) return false;
                    q.push(e.v);
                    inq[e.v] = 1;
                }
            }
        }
    }
    return true;
}

int main()
{
    fast;
    int T, kase = 1;
    cin >> T;
    while (T--)
    {
        cin >> n >> m >> x;
        for(int i = 1; i <= n; i++)
        {
            g[i].clear();
            dis[i] = 1e18;
            cnt[i] = inq[i] = 0;
        }
        for(int i = 1; i < n; i++)
        {
            addedge(i+1, i, -1);
            addedge(i, i+1, 2e9);
        }
        for(int i = 0; i < m; i++)
        {
            int a, b, c, d;
            cin >> a >> b >> c >> d;
            if(a == b)
            {
                if(c == d)
                {
                    addedge(a, c, x);
                    addedge(c, a, -x);
                }
                else
                {
                    addedge(a, c, x - 1);
                    addedge(d, a, -x - 1);
                }
            }
            else
            {
                if(c == d)
                {
                    addedge(b, c, x - 1);
                    addedge(c, a, -x - 1);
                }
                else
                {
                    addedge(b, c, x - 1);
                    addedge(d, a, -x - 1);
                }
            }
        }
        cout << "Case #" << (kase++);
        if(!spfa()) cout << " IMPOSSIBLE
";
        else {
            for(int i = 2; i <= n; i++)
                cout << ' ' << dis[i] - dis[i - 1];
            cout << endl;
        }
    }
    return 0;
}

Problem G - Alice’s Stamps 02：24：14 （-2） Solved by Dancepted （dp）

上机的时候思路有乱，调出来之后交上去因为N开大了MLE返回CE，贡献了两发罚时。

dfs枚举当前搜索过的最右端点和剩下的k的数量，记忆化一下，状态数最多就$O(n^{2})$。

HDU的T组数据比较玄学，差点没敢交。

代码：$O(n^{2})$

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 2005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x)
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

struct Node{
    int l, r;
    bool operator < (const Node& x) const {
        if (r == x.r) {
            return l < x.l;
        }
        return r < x.r;
    }
};

int n, m, k;
vector<Node> ns, ns1;
int ans = 0;
int f[N][N];
int dfs(int id, int r, int cnt, int resk) {
    if (f[r][resk]) {
        return f[r][resk];
    }
    if (resk == 0) {
        return 0;
    }
    if (id >= sz(ns)) {
        return 0;
    }
    int tmp = dfs(id+1, r, cnt, resk);
    tmp = max(tmp, ns[id].r - max(ns[id].l-1, r) + dfs(id+1, ns[id].r, cnt + ns[id].r - max(ns[id].l-1, r), resk-1));
    ans = max(ans, cnt + tmp);
    return f[r][resk] = tmp;
}
bool ok[N];
int l[N], r[N];
int main() {
    int T, kase = 1;
    cin >> T;
    while (T--) {
        ns.clear(), ns1.clear();
        read(n, m, k);
        for (int i = 1; i <= m; i++) {
            read(l[i], r[i]);
            ok[i] = true;
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= m; j++) if (i != j) {
                if (l[i] < l[j] && r[j] <= r[i]) {
                    ok[j] = false;
                }
            }
        }
        for (int i = 1; i <= m; i++) if (ok[i]) {
            ns.push_back(Node{l[i], r[i]});
        }
        sort(ns.begin(), ns.end());


        // init();
        ans = 0;
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                f[i][j] = 0;
            }
        }
        dfs(0, 0, 0, k);
        printf("Case #%d: %d
", kase++, ans);
    }
    return 0;
}

题解的做法是一个很好写的dp：

$f_{i, j}$表示最右边的覆盖点是i的情况下，用了j个集合的最大覆盖长度。

代码：$O(n^{2})$

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 2005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

int mxr[N];
int f[N][N];
int main() {
    int T; cin >> T;
    for (int kase = 1; kase <= T; kase++) {
        int n, m, k; read(n, m, k);
        memset(mxr, 0, (n+1) * sizeof(int));
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                f[i][j] = 0;
            }
        }
        for (int i = 1; i <= m; i++) {
            int l, r; read(l, r);
            mxr[l] = max(mxr[l], r);
        }
        int ans = 0, r = 0;
        for (int i = 1; i <= n; i++) {
            r = max(mxr[i], r);
            for (int j = 0; j <= k; j++) {
                f[i][j] = max(f[i][j], f[i-1][j]);
                ans = max(ans, f[i][j]);
                if (j+1 <= k)
                    f[r][j+1] = max(f[r][j+1], f[i-1][j] + r - i + 1),
                    ans = max(ans, f[r][j+1]);
            }
        }
        printf("Case #%d: %d
", kase, ans);
    }
    return 0;
}

Problem I - Inkopolis 04：59：59 （-3） Solved by Dancepted & xk

中午还在看的基环树下午就用到了。

先考虑树上的情况。会影响答案的部分只有修改颜色之前的颜色$color_{pre}$，和之后的颜色$color_{now}$两种。看是否会新增、删除color region，或者合并、分割原来的color region。

再考虑环上的情况。同上。特别地：如果整个环都是同一种颜色，修改了颜色不会分割原来的color region；如果整个环除了修改的边都是$color_{now}$，则修改颜色不会合并color region。

计算节日开始前各街道的颜色，可以模仿修改的操作，一条条边加入就行了。

实现的时候先把环扣出来，统计一下环各颜色的边数量。用个map保存每个点相邻的各颜色的数量。再用map保存每条边的颜色。

然后扫一遍所有的边先计算初始情况下的region的数量sum，然后一次次修改边的颜色，更新sum就好了。

代码：O(nlogn)

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 200005
#define M 200005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar('-'), x = -x;
    do{++len; c[len] = x%10 + '0';} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

int tot;
int head[N], nxt[N<<1], ver[N<<1], col[N<<1];
void addEdge(int u, int v, int c) {
    nxt[tot] = head[u], ver[tot] = v, col[tot] = c, head[u] = tot++;
}

int iscir[N], cirlen;
bool vis[N];
int getCir(int u, int fa, int& rt) {
    vis[u] = true;
    for (int i = head[u]; i != -1; i = nxt[i]) {
        int v = ver[i];
        if (v == fa) continue;
        if (vis[v]) {
            rt = v;
            iscir[u] = 1;
            cirlen++;
            return 1;
        }
        else {
            int tmpcir = getCir(v, u, rt);
            if (tmpcir) {
                iscir[u] = tmpcir;
                cirlen++;
                return u != rt;
            }
        }
    }
    return 0;
}

int cnt[N];
int sum;
map<int, int> mp[N];
map<pair<int,int>, int> uv_col;

void update(int u, int v, int ccur, bool ifprint) {
    if (u > v) swap(u, v);
    int cpre = uv_col[mp(u, v)];
    if (cpre == ccur) {
        if (ifprint)
            printf("%d
", sum);
        return;
    }
    if (!iscir[u] || !iscir[v]) {
        //on tree
        if (mp[u][cpre] >= 2 && mp[v][cpre] >= 2) {
            sum++;
        }
        if (mp[u][cpre] == 1 && mp[v][cpre] == 1) {
            sum--;
        }
        if (mp[u][ccur] && mp[v][ccur]) {
            sum--;
        }
        if (!mp[u][ccur] && !mp[v][ccur]) {
            sum++;
        }
    }
    else {
        //on circle
        if (mp[u][cpre] >= 2 && mp[v][cpre] >= 2) {
            if (cnt[cpre] != cirlen)
                sum++;
        }
        if (mp[u][cpre] == 1 && mp[v][cpre] == 1) {
            sum--;
        }
        if (mp[u][ccur] && mp[v][ccur]) {
            if (cnt[ccur] != cirlen-1)
                sum--;
        }
        if (!mp[u][ccur] && !mp[v][ccur]) {
            sum++;
        }

        if (cpre != -1)
            cnt[cpre]--;
        cnt[ccur]++;
    }
    uv_col[mp(u, v)] = ccur;
    if (cpre != -1) {
        mp[u][cpre]--;
        mp[v][cpre]--;
    }
    mp[u][ccur]++;
    mp[v][ccur]++;
    if (ifprint)
        printf("%d
", sum);
}

int main() {
    int T;
    cin >> T;
    for (int kase = 1; kase <= T; kase++) {
        int n, m; read(n, m);
        // init
        uv_col.clear();
        for (int i = 1; i <= n; i++) {
            mp[i].clear();
        }
        tot = 0;
        memset(head, -1, (n+1) * sizeof(int));

        // input and count colors on every edge
        for (int i = 1; i <= n; i++) {
            int u, v, c; read(u, v, c);
            addEdge(u, v, c);
            addEdge(v, u, c);
            if (u > v)
                swap(u, v);
            uv_col[mp(u, v)] = -1;
            // mp[u][c]++; mp[v][c]++;
        }
        // get circle
        int rt = -1;
        memset(vis, false, (n+1) * sizeof(bool));
        memset(iscir, 0, (n+1) * sizeof(int));
        cirlen = 0;
        getCir(1, -1, rt);
        // count colors on circle
        memset(cnt, 0, (n+1) * sizeof(int));
        sum = 0;
        for (int i = 0; i < n; i++) {
            int u = ver[i<<1], v = ver[i<<1|1], ccur = col[i<<1];
            update(u, v, ccur, false);
        }

        printf("Case #%d:
", kase);
        for (int i = 1; i <= m; i++) {
            int u, v, ccur; read(u, v, ccur);
            update(u, v, ccur, true);
        }
    }
    return 0;
}
/*
11111
5 10
1 5 1
2 5 1
3 5 1
4 5 1
1 2 1
1 2 1


2
4 4
1 2 1
2 3 1
3 4 1
4 1 1
1 2 2
3 4 2
2 3 2
4 1 4
4 3
4 2 4
2 3 3
3 4 2
1 4 1
3 4 2
2 3 4
3 4 3
*/

---

总结：

本场能绝杀还是有点运气成分在的，而且现场赛的话最后几分钟可能就交不上题了也可能。不过绝杀也不影响苟在银牌，问题不大。

然后最近的几场好像都是我在贡献罚时，感觉码力出现了一些小小的问题。这两天要小心一点。

然后还有两周可能就要退役了，模拟赛还没有进过金区感觉很难受啊，打成这个样子还怎么冲金啊qaq。