LOJ刷题记录：2024-2029（SHOI2016）

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loj#2024. 「JLOI / SHOI2016」侦查守卫

设f[nd][k]为nd这个子树，至少覆盖到包括nd上面k层；若k<0，则表示不包括nd，下面−k层已经被覆盖。然后讨论两种转移情况：

1. 这个节点放一个
2. 这个节点不放

要分开判断是欠着还是富余…

然后就是码农题了..

#include <bits/stdc++.h>using namespace std;const int MAXN = 500005;struct node {    int to, next;} edge[MAXN*2];int head[MAXN], top = 0;inline void push(int i, int j){ edge[++top] = (node) {j, head[i]}, head[i] = top;}int f[MAXN][45]; // f[nd][k] --> nd这个子树，至少覆盖到包括nd上面k层；若k<0，则表示不包括nd，下面-k层已经被覆盖int w[MAXN], hav[MAXN];int n, m, d;void dfs(int nd, int fa){    int sd[45];    memset(sd, 0, sizeof sd);    for (register int i = head[nd]; i; i = edge[i].next) {        if (edge[i].to == fa) continue;            dfs(edge[i].to, nd);        for (int j = -d; j <= d+1; j++)            sd[j+22] += f[edge[i].to][j+22];    }    for (register int k = -d; k <= d+1; k++) {        f[nd][k+22] = w[nd];        for (register int i = head[nd]; i; i = edge[i].next) {            if (edge[i].to == fa) continue;            f[nd][k+22] += f[edge[i].to][-d+1+22];        }        if (k <= 0) f[nd][k+22] = min(f[nd][k+22], sd[k+1+22]);        if (k == 1 && !hav[nd]) f[nd][k+22] = min(f[nd][k+22], sd[1+22]);        int need = k;        if (k != 1 || hav[nd]) need++;        if (need < 1) need = 1;        for (register int i = head[nd]; i; i = edge[i].next) { // 有干扰，枚举干扰节点            if (edge[i].to == fa) continue;            for (register int j = max(need, 2); j <= d+1; j++) {                f[nd][k+22] = min(f[nd][k+22], f[edge[i].to][j+22]+sd[3-j+22]-f[edge[i].to][3-j+22]);            }         }    }}int main(){    memset(f, -1, sizeof f);    scanf("%d%d", &n, &d);    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);    scanf("%d", &m);    for (register int i = 1; i <= m; i++) {        int u; scanf("%d", &u);        hav[u] = 1;    }    for (register int i = 1; i < n; i++) {        int u, v; scanf("%d%d", &u, &v);        push(u, v), push(v, u);    }    dfs(1, 0);    printf("%d\n", f[1][1+22]);    return 0;}

loj#2027. 「SHOI2016」黑暗前的幻想乡

SH·数数大赛·OI….

枚举子集，然后用矩阵树算一算…再容斥起来。复杂度及其不科学..然而跑得比谁都快

#include <bits/stdc++.h>using namespace std;const int MAXN = 20, mod = 1e9+7;int n;vector<pair<int, int> > M[MAXN];int g[MAXN][MAXN];inline int power(int a, int n){    int ans = 1;    for (int i = 0; i <= 30; i++) {        if (n&(1<<i)) ans = (long long)ans*a%mod;        a = (long long)a*a%mod;    }    return ans;}inline int inv(int a){ return power(a, mod-2); }int gauss(){    int flag = 1;    for (int i = 1; i < n; i++) {        int pos = i;        while (pos < n && !g[pos][i]) pos++;        if (pos == n) return 0;        if (pos != i) swap(g[pos], g[i]), flag *= -1;        for (int j = 1; j < n; j++) {            if (j == i) continue;            int tmp = ((-(long long)g[j][i]*inv(g[i][i]))%mod+mod)%mod;            for (int k = 1; k < n; k++)                g[j][k] = (g[j][k]+(long long)g[i][k]*tmp)%mod;        }    }    for (int i = 1; i < n; i++)        flag = (long long)flag*g[i][i]%mod;    return (flag+mod)%mod;}int main(){    scanf("%d", &n);    for (int i = 1; i < n; i++) {        int mi, u, v; scanf("%d", &mi);        for (int j = 1; j <= mi; j++) {            scanf("%d%d", &u, &v);            M[i].push_back(make_pair(u, v));        }    }    int ans = 0;    for (int i = 0; i < (1<<(n-1)); i++) {        int num = 0;        memset(g, 0, sizeof g);        for (int j = 1; j < n; j++) {            if (!(i&(1<<(j-1)))) continue;            num++;            for (int k = 0; k < M[j].size(); k++) {                int u = M[j][k].first, v = M[j][k].second;                g[u][v]--, g[v][u]--, g[u][u]++, g[v][v]++;            }        }        if ((n-1-num)&1) ans -= gauss();        else ans += gauss();        ((ans %= mod) += mod) %= mod;    }    printf("%d\n", ans);    return 0;}

loj#2028. 「SHOI2016」随机序列

每一个加号都会有一个减号消掉他，所以贡献只来源于第一个加减前面的..所以直接维护就好了。

#include <bits/stdc++.h>using namespace std;const int MAXN = 100005, mod = 1e9+7;int n, q;int arr[MAXN];int a[MAXN*4];int lc[MAXN*4], rc[MAXN*4], l[MAXN*4], r[MAXN*4], root, top = 0, dat[MAXN*4];int tag[MAXN*4];int power(int a, int n){    int ans = 1;    for (int i = 0; i <= 30; i++) {        if (n&(1<<i)) ans = (long long)ans*a%mod;        a = (long long)a*a%mod;    }    return ans;}int inv(int a){ return power(a, mod-2); }void build(int &nd, int opl, int opr){    nd = ++top, l[nd] = opl, r[nd] = opr, tag[nd] = 1;    if (opl == opr) dat[nd] = a[opl];    else {        int mid = (l[nd]+r[nd])>>1;        build(lc[nd], opl, mid), build(rc[nd], mid+1, opr);        dat[nd] = (dat[lc[nd]]+dat[rc[nd]])%mod;    }}void pdw(int nd){    if (lc[nd]) tag[lc[nd]] = (long long)tag[nd]*tag[lc[nd]]%mod, tag[rc[nd]] = (long long)tag[nd]*tag[rc[nd]]%mod;    dat[nd] = (long long)dat[nd]*tag[nd]%mod, tag[nd] = 1;}void modify(int nd, int opl, int opr, int dt){    // cerr << nd << " " << opl << " " << opr << " " << dt << endl;    pdw(nd);    if (l[nd] == opl && r[nd] == opr) tag[nd] = (long long)tag[nd]*dt%mod;    else {        int mid = (l[nd]+r[nd])/2;        if (opr <= mid) modify(lc[nd], opl, opr, dt);        else if (opl > mid) modify(rc[nd], opl, opr, dt);        else modify(lc[nd], opl, mid, dt), modify(rc[nd], mid+1, opr, dt);        pdw(lc[nd]), pdw(rc[nd]);        dat[nd] = (dat[lc[nd]]+dat[rc[nd]])%mod;    }}int main(){    // cerr << (long long)2414*inv(2414)%mod << endl;    scanf("%d%d", &n, &q);    for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);    int mul = 1;    for (int i = 1; i <= n; i++) {        mul = (long long)mul*arr[i]%mod;        if (i < n) a[i] = (long long)mul*2*power(3, n-i-1)%mod;        else a[i] = mul;    }    build(root, 1, n);    // cerr << dat[root] << endl;    for (int i = 1; i <= q; i++) {        int t, v;        scanf("%d%d", &t, &v);        modify(root, t, n, (long long)inv(arr[t])*v%mod), arr[t] = v;        pdw(root), printf("%d\n", dat[root]);    }    return 0;}