省选测试2

A 青蛙

题目大意：给定d，如果第 i 只青蛙的某次跳跃的距离超过 d，那么需要付出 (c_i) 的代价，只有给定的石头可以跳，且只能跳一次，不能不跳，问全部青蛙过河的最小费用

首先想到先让花费大的跳给定的石头，让他们免费跳，最后没石头了花费小的再跳
发现如果有一只可以免费跳过去，那么可以保证所有的都跳过去
所以求出最多可以免费跳过 m 只青蛙，让费用最大的 m 只青蛙免费跳过并踩掉所有石头、剩下的青蛙直接从 1 跳到 n。
如果 m 为 0，那就先让费用最小的把所有石头都踩掉就好了
如果1 和 n 之间的距离小于 d，那全部青蛙都可以免费跳过了
至于如何求 m，这就是之前的一道青蛙过河的原题了
看每个从1不能到的石头有几个前面的石头可以到自己，取个min就是m

Code

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#include <cstdio>
#include <algorithm>

const int N = 1e5 + 5;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

int m, n, k, d, p[N];
long long s[N], ans;

int main() {
    for (int T = read(); T; --T) {
        m = read(), n = read(); k = read() + 2; d = read(); ans = 0;
        for (int i = 1; i <= n; ++i) s[i] = read();
        for (int i = 2; i < k; ++i) p[i] = read();
        p[1] = 1; p[k] = m;
        std::sort(s + 1, s + n + 1);
        std::sort(p + 1, p + k + 1);
        for (int i = 1; i <= n; ++i) s[i] += s[i-1];
        for (int i = 1; i <= k; ++i)
            if (p[i] > d + 1) m = std::min(m, p + i - std::lower_bound(p + 1, p + i, p[i] - d));
        if (m) ans = s[1];
        else {
            for (int i = 2; i <= k; ++i)
                if (p[i] - p[i-1] > d) ans += s[1];
        }
        printf("%lld
", 1 + d >= p[k] ? 0ll : ans + s[n-m] - s[1]);
    }
    return 0;
}

B 蚯蚓排队

题目大意：动态连接字符串，回答一个字符串的出现次数

因为 k 只有 50，每次连接断开的时候枚举每个改变的字符串在哈希表里更新就好了

Code

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#include <cstdio>
#include <cstring>

typedef unsigned long long ull;
const int N = 2e5 + 5, M = 998244353, bs = 13331;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

struct Hash {
#define M 5000000
    int h[M], n[M], t[M], hac; ull w[M];
    int & operator [] (ull ha) {
        int x = ha % M;
        for (int i = h[x]; i; i = n[i])
            if (w[i] == ha) return t[i];
        w[++hac] = ha; n[hac] = h[x]; h[x] = hac;
        return t[hac];
    }
#undef M
}mp;

int n, m, a[N], l[N], r[N], len;
char c[N*50];
ull P[55];

void Update(int x, int w) {
    ull s1 = 0, s2;
    for (int p = x, i = 0; p && i <= 48; p = l[p], ++i) {
        s1 += a[p] * P[i]; s2 = s1;
        for (int q = r[x], j = i + 2; q && j <= 50; q = r[q], ++j)
            s2 = s2 * bs + a[q], mp[s2] += w;
    }
}

int main() {
    n = read(); m = read(); P[0] = 1;
    for (int i = 1; i <= n; ++i) a[i] = read(), mp[a[i]]++;
    for (int i = 1; i <= 50; ++i) P[i] = P[i-1] * bs;
    while (m--) {
        int od = read();
        if (od == 1) {
            int x = read(), y = read();
            r[x] = y; l[y] = x; Update(x, 1);
        }
        else if (od == 2) {
            int x = read(), y = r[x];
            Update(x, -1); r[x] = l[y] = 0;
        }
        else {
            scanf("%s", c + 1); len = strlen(c + 1);
            int k = read(), ans = 1; ull s = 0;
            for (int i = 1; i <= len; ++i) {
                s = s * bs + (c[i] -= '0');
                if (i >= k) ans = 1ll * ans * mp[s-=P[k]*c[i-k]] % M;
            }
            printf("%d
", ans);
        }
    }
    return 0;
}

C 最大价值

题目大意：从n个物品里选i个使价值最大

因为如果第i个物品放在第j位，他的贡献是(j-1)*a[i]+b[i]，所以a越大放的位置越靠后，所以先按a排序
然后就可以n2的Dp了,f[i][j]=max(f[i-1][j],f[i-1][j-1]+(j-1)*a[i]+b[i])
会发现一个事情，如果第i件物品放在第j位比前面找出来的方案更优，那第i件放在第j+1到i位都可以更优
然后看差分数组，发现前面的不变，只是在第一次改变的地方j这里，差分数组插入了一个(j-1)*a[i]+b[i],然后后面的都加a[i]
于是可以有Splay维护差分数组进行DP

Code

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#include <cstdio> 
#include <algorithm>
#define Get(x) (ch[fa[x]][1] == x) 

using namespace std;
typedef long long ll;
const int N = 3e5 + 5;

ll read(ll x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar()) x = x * 10 + c - '0';
    return x * f;
}

int n, a[N], p[N], fa[N], sz[N], ch[N][2], rt, stk[N], tp;
ll b[N], w[N], tag[N], ans;

bool Cmp(int x, int y) {
    return a[x] < a[y];
}

void Pushup(int x) {
    sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
}

void Rotate(int x) {
    int y = fa[x], z = fa[y], k = Get(x), B = ch[x][k^1];
    ch[z][Get(y)] = x; fa[x] = z;
    ch[x][k^1] = y; fa[y] = x;
    ch[y][k] = B; fa[B] = y;
    Pushup(y); Pushup(x);
}

void Update(int x, ll v) {
    w[x] += v, tag[x] += v;
}

void Pushdown(int x) {
    if (!tag[x]) return;
    Update(ch[x][0], tag[x]);
    Update(ch[x][1], tag[x]);
    tag[x] = 0;
}

void Splay(int x) {
    for (int i = x; i; i = fa[i]) stk[++tp] = i;
    while (tp) Pushdown(stk[tp--]);
    while (fa[x] != 0) {
        int y = fa[x];
        if (fa[y] != 0) Get(x) != Get(y) ? Rotate(x) : Rotate(y);
        Rotate(x);
    }
    rt = x;
}

void Print(int x) {
    if (!x) return;
    Pushdown(x);
    Print(ch[x][0]);
    printf("%lld
", ans += w[x]);
    Print(ch[x][1]);
}

int main() {
    n = read();
    for (int i = 1; i <= n; ++i)
        a[i] = read(), b[i] = read(), p[i] = i;
    sort(p + 1, p + n + 1, Cmp);
    w[1] = b[p[1]]; rt = 1;
    for (int i = 2; i <= n; ++i) {
        int x = rt, s = 0, lt, k;
        while (x) {
            Pushdown(lt = x);
            if (w[x] < 1ll * (s + sz[ch[x][0]]) * a[p[i]] + b[p[i]]) k = 0;
            else k = 1, s += sz[ch[x][0]] + 1;
            x = ch[lt][k];
        }
        ch[lt][k] = i; fa[i] = lt; w[i] = 1ll * s * a[p[i]] + b[p[i]];
        Splay(i); Update(ch[i][1], a[p[i]]);
    }
    Print(rt); return 0;
}