Codeforces Round #344 (Div. 2)

 

水 A - Interview

注意是或不是异或

#include <bits/stdc++.h>

int a[1005], b[1005];

int main()  {
    int n;  scanf ("%d", &n);
    for (int i=0; i<n; ++i) {
        scanf ("%d", a+i);
    }
    for (int i=0; i<n; ++i) {
        scanf ("%d", b+i);
    }
    int ans = 0;
    for (int i=0; i<n; ++i) {
        int x = 0, y = 0;
        for (int j=i; j<n; ++j) {
            x |= a[j];  y |= b[j];
            ans = std::max (ans, x + y);
        }
    }
    printf ("%d\n", ans);

    return 0;
}

水 B - Print Check

color[i][j] = max (timei(rowi), timej(colj)))

#include <bits/stdc++.h>

const int N = 5e3 + 5;
int a[N][N];
std::pair<int, int> row[N], col[N];

int main() {
    int n, m, k;    scanf ("%d%d%d", &n, &m, &k);
    for (int op, x, y, i=1; i<=k; ++i) {
        scanf ("%d%d%d", &op, &x, &y);
        if (op == 1) {
            row[x] = std::make_pair (y, i);
        } else {
            col[x] = std::make_pair (y, i);
        }
    }
    for (int i=1; i<=n; ++i) {
        for (int j=1; j<=m; ++j) {
            int tr = row[i].second, tc = col[j].second;
            if (tr > tc && tr > 0) {
                a[i][j] = row[i].first;
            } else if (tc > tr && tc > 0) {
                a[i][j] = col[j].first;
            }
        }
    }
    for (int i=1; i<=n; ++i) {
        for (int j=1; j<=m; ++j) {
            printf ("%d%c", a[i][j], j == m ? '\n' : ' ');
        }
    }

    return 0;
}

单调队列+排序 C - Report

题意：给m次操作，每次操作使得[1, r]范围的的a[i]升序或降序排序，问a[i]最后的结果

分析：因为左端点固定，右端点远的且操作时间靠后的可以覆盖之前右端点近的操作，所以先从右端点最远且时间靠后的时间出发，之后类似只要处理右端点降序的位置，即维护一个单调队列。队列中后一次操作长度不大于前一次长度，减少的长度k里的数字为前一次范围中前k大的数或前k小的数。

#include <bits/stdc++.h>

const int N = 2e5 + 5;
int a[N], b[N];
int r[N], t[N];

int main() {
    int n, m; scanf ("%d%d", &n, &m);
    for (int i=1; i<=n; ++i) {
        scanf ("%d", a+i);
    }
    int s = 0;
    for (int x, y, i=1; i<=m; ++i) {
        scanf ("%d%d", &x, &y);
        while (s > 0 && y >= r[s-1])    s--;
        t[s] = x;   r[s] = y;   s++;
    }
    r[s++] = 0;

    int bl = 1, br = r[0];
    for (int i=bl; i<=br; ++i) b[i] = a[i];
    std::sort (b+1, b+1+br);

    for (int i=1; i<s; ++i) {
        for (int j=r[i-1]; j>r[i]; --j) {
            a[j] = (t[i-1] == 1) ? b[br--] : b[bl++];
        }
    }

    for (int i=1; i<=n; ++i) {
        printf ("%d ", a[i]);
    }
    puts ("");

    return 0;
}

KMP D - Messenger

题意：给两个字符串，问后者在前者里出现的次数。给出的方式：1-a 2-b 3-c 4-d(abbcccddd)

分析：其实想明白就知道这就是简单的KMP问题，文本串第一个和最后一个的个数一定要比模式串多，且里面的要完全相等。还要考虑特殊情况，压缩后m==1或2时。

#include <bits/stdc++.h>

typedef long long ll;
const int N = 2e5 + 5;
struct Part {
    ll len; char ch;
    bool operator == (const Part &rhs) const {
        return len == rhs.len && ch == rhs.ch;
    }
    bool operator >= (const Part &rhs) const {
        return len >= rhs.len && ch == rhs.ch;
    }
    bool operator != (const Part &rhs) const {
        return len != rhs.len || ch != rhs.ch;
    }
};
Part A[N], B[N];
int fail[N];

int compress(Part *C, int n) {
    int m = 0;
    for (int i=1; i<n; ++i) {
        if (C[m].ch == C[i].ch) {
            C[m].len += C[i].len;
        } else {
            C[++m] = C[i];
        }
    }
    return m + 1;
}

void get_fail(Part *P, int lenp) {
    int i = 0, j = -1;  fail[0] = -1;
    while (i < lenp) {
        if (j == -1 || P[j] == P[i]) {
            ++i; ++j;   fail[i] = j;
        } else {
            j = fail[j];
        }
    }
}

ll KMP(Part *C, int n, Part *D, int m) {
    get_fail (D, m);
    int i = 0, j = 0;
    ll ret = 0;
    while (i < n) {
        while (j != -1 && C[i] != D[j]) j = fail[j];
        i++;    j++;
        if (j == m) {
            if (C[i] >= D[m] && C[i-m-1] >= D[-1]) ret++;
            j = fail[j];
        }
    }
    return ret;
}

int main() {
    int n, m;   scanf ("%d%d", &n, &m);
    int bug = 0;
    char str[3];
    for (int i=0; i<n; ++i) {
        scanf ("%I64d-%s", &A[i].len, str);
        A[i].ch = str[0];
    }
    for (int i=0; i<m; ++i) {
        scanf ("%I64d-%s", &B[i].len, str);
        B[i].ch = str[0];
    }
    n = compress (A, n);
    m = compress (B, m);
    ll ans = 0;
    if (m == 1) {
        for (int i=0; i<n; ++i) {
            if (A[i].ch == B[0].ch) {
                ans += std::max (0ll, A[i].len - B[0].len + 1);
            }
        }
    } else if (m == 2) {
        for (int i=0; i<n-1; ++i) {
            if (A[i] >= B[0] && A[i+1] >= B[1]) ans++;
        }
    } else {
        ans = KMP (A+1, n-2, B+1, m-2);
    }
    printf ("%I64d\n", ans);

    return 0;
}

凸包 E - Product Sum

题意：将一个点移动到另一个点(可相同)，求

分析：推算化简的得到： 

　　　　　　　　或者 

先考虑后者，再次化简得到：，，可以抽象成ax+b, (x = ar),所以我们枚举ar，得到最大的，用到了凸包，把直线加入凸包，得到单调上升的类似于这样的，然后二分找最大值。前者类似。

 

第一次碰到这种题，写详细点:)

#include <bits/stdc++.h>

typedef long long ll;
const int N = 2e5 + 5;
int a[N];
ll sum[N];

struct Line {
    int a; ll b;
    ll get(int x) {
        return 1ll * a * x + b;
    }
};
struct Convex_hull {
    int sz;
    Line *hull;
    Convex_hull(int maxn) {
        hull = new Line[++maxn]; sz = 0;
    }
    void clear() {
        sz = 0;
    }
    bool is_bad(int cur, int pre, int nex) {
        Line c = hull[cur], p = hull[pre], n = hull[nex];
        return (c.b - n.b) * (c.a - p.a) <= (p.b - c.b) * (n.a - c.a);
    }
    void add_line(int a, ll b) {
        hull[sz++] = (Line) {a, b};
        while (sz > 2 && is_bad (sz-2, sz-3, sz-1)) {
            hull[sz-2] = hull[sz-1]; sz--;
        }
    }
    ll query(int x) {
        int low = -1, high = sz - 1;
        while (low + 1 < high) {
            int mid = low + high >> 1;
            if (hull[mid].get (x) <= hull[mid+1].get (x)) {
                low = mid;
            } else {
                high = mid;
            }
        }
        return hull[high].get (x);
    }
};

int main() {
    int n; scanf ("%d", &n);
    ll ans = 0; sum[0] = 0;
    for (int i=1; i<=n; ++i) {
        scanf ("%d", a+i);
        sum[i] = sum[i-1] + a[i];
        ans += 1ll * a[i] * i;
    }
    Convex_hull *hull = new Convex_hull (n);
    ll add = 0;
    for (int r=2; r<=n; ++r) {
        hull->add_line (r-1, -sum[r-2]);
        add = std::max (add, hull->query (a[r]) + sum[r-1] - 1ll * a[r] * r);
    }
    hull->clear ();
    for (int l=n-1; l>=1; --l) {
        hull->add_line (-(l+1), -sum[l+1]);
        add = std::max (add, hull->query (-a[l]) + sum[l] - 1ll * a[l] * l);
    }
    printf ("%I64d\n", ans + add);

    return 0;
}