2020 2.4

t1

给出一个图，求图中所有不同有根树的权值之和。

定义有根树权值为

 d表示点的深度，根的深度为0.

n<=10 m<=1000

题解

还是类似于宝藏的状压。f[t][i][j]表示叶子距离根节点距离为t，非叶子节点为状态i,叶子节点为状态j，权值的和，g[t][i][j]表示方案数。

f[t][i][j]就可以转移到f[t+1][i|j][k](k和i|j不相交)，需要预处理出叶节点为j转移到叶节点为k的权值和，方案数。

代码如下：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n, m, x[2020], y[2020], w[2020], p = 1000000007;
int f[11][1025][1025]; // cnt
int g[11][1025][1025]; // sum
ll c[1025][1025];
ll s[1025][1025];
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &x[i], &y[i], &w[i]);
        x[i]--;
        y[i]--;
        x[i + m] = y[i];
        y[i + m] = x[i];
        w[i + m] = w[i];
    }
    m *= 2;
    for (int i = 0; i < 1 << n; i++) {
        for (int j = 0; j < 1 << n; j++) {
            if (i & j) {
                continue;
            }
            ll ts[11] = {};
            ll tc[11] = {};
            for (int k = 0; k < m; k++) {
                if ((i >> x[k] & 1) && (j >> y[k] & 1)) {
                    tc[y[k]]++;
                    ts[y[k]] += w[k];
                }
            }
            c[i][j] = 1;
            s[i][j] = 0;
            for (int k = 0; k < n; k++) {
                if (j >> k & 1) {
                    c[i][j] *= tc[k];
                }
            }
            if (c[i][j] == 0) {
                continue;
            }
            for (int k = 0; k < n; k++) {
                if (j >> k & 1) {
                    s[i][j] += c[i][j] / tc[k] * ts[k];
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        f[0][0][1 << i] = 1;
        g[0][0][1 << i] = 0;
    }
    for (int t = 0; t < n; t++) {
        for (int i = 0; i < 1 << n; i++) {
            for (int j = 0; j < 1 << n; j++) {
                if (i & j) {
                    continue;
                }
                if (f[t][i][j] == 0) {
                    continue;
                }
                int mask = (1 << n) - 1 - i - j;
                for (int k = mask;; k = (k - 1) & mask) {
                    ll gc = (ll)g[t][i][j] * c[j][k];
                    ll fc = (ll)f[t][i][j] * c[j][k];
                    ll fs = (ll)f[t][i][j] * s[j][k] * (t + 1);
                    f[t + 1][i | j][k] = (f[t + 1][i | j][k] + fc) % p;
                    g[t + 1][i | j][k] = (g[t + 1][i | j][k] + gc + fs) % p;
                    if (k == 0) {
                        break;
                    }
                }
            }
        }
    }
    printf("%d
", g[n][(1 << n) - 1][0]);
    return 0;
}

t2 略

t3

4 个字符串，求LCS。每个字符串中，前三个每个数字⾄多出现2 次。

n<=10000.

可以记录前三个字符串每个数字出线的位置，在第四个字符串找到数d[i]时，在前三个里各取一个位置a[i].b[i],c[i]来与d[i]构成一个点。因为前面每个最多出现两次，所以最多8n个点，跑四位正序。

代码如下：

#include <bits/stdc++.h>

#define MAXN 200005

using namespace std;

int n, BIT[MAXN], ans, turn;

struct Data{
    int id, a, b, c;
    bool type; // false change / true query
}op[4][MAXN];

int f[MAXN];
bool cmp(const Data &A, const Data &B) {
    if (turn == 1) {
        if (A.a != B.a) {
            return A.a < B.a;
        }
        if (A.b != B.b) {
            return A.b > B.b;
        }
        return A.c > B.c;
    }
    if (turn == 2A) {
        if (A.b != B.b) {
            return A.b < B.b;
        }
        return A.c > B.c;
    }
    assert(false);
}

int lowbit(int x){
    return x & (-x);
}

void clear(int x) {
    x++;
    for (int i = x; i <= MAXN; i += i & -i) {
        BIT[i] = 0;
    }
}

void change(int x, int v) {
    x++;
    for (int i = x; i <= MAXN; i += i & -i) {
        BIT[i] = max(BIT[i], v);
    }
}

int query(int x) {
    x++;
    int re = 0;
    for (int i = x; i > 0; i -= i & -i) {
        re = max(re, BIT[i]);
    }
    return re;
}

void Work(int cnt) {
//    printf("start
");
    for (int i = 0; i < cnt; i++) {
//        printf("d%d %d %d %d %d
", op[2][i].id, op[2][i].a, op[2][i].b, op[2][i].c, op[2][i].type);
        if (!op[2][i].type) {
            change(op[2][i].c, f[op[2][i].id]);
        } else {
            f[op[2][i].id] = max(f[op[2][i].id], query(op[2][i].c - 1) + 1);
        }
    }
    for (int i = 0; i < cnt; i++) {
        if (!op[2][i].type) {
            clear(op[2][i].c);
        }
    }
//    printf("end
");
}

void CDQ(int now, int L, int R) {
    if (R - L <= 1) {
        return;
    }
    int mid = (L + R) / 2;
    int cnt = 0;
    CDQ(now, L, mid);

    for (int i = L; i < mid; i++) {  
        if (now == 1) {
            op[now][cnt] = op[now - 1][i];
            op[now][cnt].type = false;
            cnt++;
        } else if (!op[now - 1][i].type) {
            op[now][cnt] = op[now - 1][i];
            cnt++;
        }
    }
    for (int i = mid; i < R; i++) {
        if (now == 1) {
            op[now][cnt] = op[now - 1][i];
            op[now][cnt].type = true;
            cnt++;
        } else if (op[now - 1][i].type) {
            op[now][cnt] = op[now - 1][i];
            cnt++;
        }
    }

    turn = now;
    sort(op[now], op[now] + cnt, cmp);
    if(now == 2) {
//        printf("%d %d %d %d
", now, L, R, cnt);
        Work(cnt);
    } else {
        CDQ(now + 1, 0, cnt);
    }
    CDQ(now, mid, R);
}


int A[50020];
int B[50020];
int C[50020];
int D[50020];
vector<int> pA[50020];
vector<int> pB[50020];
vector<int> pC[50020];
void read() {
    int N;
    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf("%d", &A[i]);
        pA[A[i]].push_back(i);
    }
    for (int i = 0; i < N; i++) {
        scanf("%d", &B[i]);
        pB[B[i]].push_back(i);
    }
    for (int i = 0; i < N; i++) {
        scanf("%d", &C[i]);
        pC[C[i]].push_back(i);
    }
    for (int i = 0; i < N; i++) {
        scanf("%d", &D[i]);
        for (int ja = 0; ja < pA[D[i]].size(); ja++) {
            int pa = pA[D[i]][ja];
            for (int jb = 0; jb < pB[D[i]].size(); jb++) {
                int pb = pB[D[i]][jb];
                for (int jc = 0; jc < pC[D[i]].size(); jc++) {
                    int pc = pC[D[i]][jc];
                    op[0][n] = {n, pa, pb, pc, false};
                    n++;
                }
            }
        }
    }

}
int main(){
    read();
    for (int i = 0; i < n; i++) {
//        printf("%d %d %d %d
", op[0][i].id, op[0][i].a, op[0][i].b, op[0][i].c);
    }
    for (int i = 0; i < n; i++) {
        f[i] = 1;
    }
    CDQ(1, 0, n);
    int ans = 0;
    for (int i = 0; i < n; i++) {
        ans = max(ans, f[i]);
    }
//    fprintf(stderr, "n = %d
", n);
    printf("%d
", ans);
    return 0;
}