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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#620400#2438. Minimum Spanning TreesAfterlife#AC ✓1372ms5016kbC++205.2kb2024-10-07 17:57:142024-10-07 17:57:14

Judging History

你现在查看的是最新测评结果

  • [2024-10-07 17:57:14]
  • 评测
  • 测评结果:AC
  • 用时:1372ms
  • 内存:5016kb
  • [2024-10-07 17:57:14]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n , k;
int p[6];
int s[6];
int fpow(int a,int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a %mod;
        a = 1LL * a * a  %mod ; b >>= 1;
    }
    return ans;
}
typedef vector<int> poly;
poly operator * (const poly &a,const poly& b) {
    // assert(a.size() && b.size()) ;
    if(!a.size() || (!b.size())) {
        poly c ; return c;
    }
    poly c((int)a.size() + b.size() - 1);
    for(int i = 0;i < a.size();i++) {
        for(int j = 0;j < b.size();j++) {
            c[i + j] = (c[i + j] + 1LL * a[i] * b[j]) % mod;
        }
    }
    return c;
}
poly operator + (const poly &a,const poly& b) {
    poly c(max(a.size() , b.size()));
    for(int i = 0;i < a.size();i++) c[i] = a[i];
    for(int i = 0;i < b.size();i++) c[i] = (c[i] + b[i]) % mod;
    return c;
}
poly operator * (const poly &a,int b) {
    poly c(a);
    for(int i = 0;i < c.size();i++) c[i] = 1LL * c[i] * b % mod;
    return c;
}
int t[45] , rt[45];

poly h[5][41];
poly f[5][41][41];
poly g[41][41];

poly mul(const poly& a,int d) {
    poly c(a.size() + d) ;
    for(int i = 0;i < a.size();i++) c[i + d] = a[i];
    return c;
}
void solv() {
    cin >> n>> k;
    int r100 = fpow(100 , mod - 2);
    for(int i = 0;i <= k;i++) {
        cin >> p[i]; p[i] = 1LL * p[i] * r100 % mod;
    }
    s[k + 1] = p[0];
    for(int i = k;i >= 0;i--) s[i] = (s[i + 1] + p[i]) % mod;
    t[0] = rt[0] = 1;
    for(int i = 1;i <= n;i++) t[i] = 1LL * t[i - 1] * i % mod , rt[i] = fpow(t[i] , mod-2);
    h[0][1] = vector<int>(1 , 1) ;

    for(int i = 1;i <= k;i++) {
        vector<vector<int>> p0(n+1 , vector<int>(n+1)) , p1(n+1 , vector<int>(n+1)) , p2(n + 1 , vector<int>(n + 1));
        for(int x = 0 ; x <= n;x++) {
            for(int y = 0;y <= n;y++) {
                p1[x][y] = (fpow(s[i] , x * y) - fpow(s[i + 1] , x * y) + mod) % mod;
                p0[x][y] = fpow(s[i + 1] , x * y) ;
                p2[x][y] = (p0[x][y] + p1[x][y]) % mod;
                // printf("%d %d : %d %d\n",x,y,p0[x][y],p1[x][y]);
            }
        }
        for(int s = 1 ; s <= n;s++) {
            for(int x = 1 ; x <= s ; x++) {
                int y = s - x; f[i][x][y].clear() ;
            }
        }
        for(int x = 1 ; x <= n;x++) f[i][x][0] = h[i - 1][x] * rt[x - 1] ;
        for(int s = 1 ; s <= n;s++) {
            for(int x = 1;x <= s;x++) {
                int y = s - x;
                if(!f[i][x][y].size()) continue ;
                int lft = n - s;
                for(int mx = 0; mx <= lft;mx++) for(int u = 0 ; u <= lft;u++) g[mx][u].clear() ;
                g[0][0] = vector<int>(1 , 1);
                for(int mx = 0 ; mx < lft ; mx++) { /// mx
                    for(int u = 0 ; u <= lft ; u++) { 
                        if(!g[mx][u].size()) continue ;
                        // printf("%d %d\n" , mx , u) ;
                        poly L = mul(h[i - 1][mx + 1] * rt[mx + 1] * p0[mx + 1][y] * p1[mx + 1][x] , i - 1);
                        poly Lk(1 , 1);
                        // if(mx == 1 && u == 0 && i == 2 && x == 1 && y == 0) {
                        //     printf("L : ");
                        //     for(auto x : L) printf("%d ",x) ;
                        //     printf("\n");
                        // }
                        for(int j = 0 ; j * (mx + 1) + u <= lft ; j++) {
                            g[mx + 1][u + j * (mx + 1)] = g[mx + 1][u + j*(mx + 1)] + g[mx][u] * Lk * rt[j] ;
                            // printf("J %d : %d %d\n",j,mx + 1, u + j * (mx + 1));
                            if((j + 1) * (mx + 1) + u <= lft) {Lk = Lk * L * p2[mx + 1][u + j * (mx + 1)]; }
                        }
                        // puts("OK") ;
                        // for(int d = max(1,mx) ; d + u <= lft ; d++) {
                        //     /// to g[d + u][d]
                        //     g[d + u][d] = g[d + u][d] + mul(g[u][mx] * h[i - 1][d] * rt[d] * p0[d][y] * p1[d][x] , i - 1) ;
                        // }
                    }
                }
                // printf("%d %d : ",x,y);
                // for(auto x : f[i][x][y]) printf("%d ",x) ; printf("\n");
                    
                
                // vector<poly> t(n + 1) ;
                for(int u = 1 ; u <= lft ; u++) {
                    // for(int i = 1;i <= u;i++) {
                    //     t[u] = t[u] + g[i][u] ;
                    // }
                    // printf("goto %d : ",u);
                    // for(auto x : g[u][u]) printf("%d ",x) ; printf("\n");
                
                    f[i][u][x + y] = f[i][u][x + y] + f[i][x][y] * g[u][u];
                }
            }
        }
        
        for(int x = 1;x <= n;x++) {
            h[i][x].clear() ;
            for(int a = 1 ; a <= x;a++) {
                h[i][x] = h[i][x] + f[i][a][x - a] * t[x - 1];
            }
        }
    }
    for(int s = 0 ; s <= (k - 1) * (n - 1) ; s++) {
        cout << h[k][n][s] <<' ' ;
    }
    cout << '\n';
    return ;
}
int main() {
    ios::sync_with_stdio(false) ; cin.tie(0);
    int t;cin >> t;
    while(t--) solv() ;
}

Details

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Test #1:

score: 100
Accepted
time: 1372ms
memory: 5016kb

input:

200
3 1
50 50
3 2
0 50 50
3 3
25 25 25 25
8 1
41 59
7 3
37 30 7 26
3 3
16 12 18 54
9 2
9 43 48
9 3
3 40 42 15
9 1
29 71
9 2
40 42 18
5 1
76 24
5 1
39 61
9 2
23 38 39
10 4
18 15 34 2 31
7 2
23 28 49
9 4
15 13 25 19 28
7 1
64 36
6 1
50 50
9 1
4 96
4 1
64 36
9 2
24 45 31
9 2
3 61 36
9 1
65 35
8 4
6 1 3...

output:

500000004 
500000004 375000003 125000001 
406250003 109375001 250000002 265625002 562500004 
858129220 
40267248 73443306 307645653 13908396 542571454 781149891 223877799 478284083 469782292 483514097 271207900 851118600 686534546 
708608005 271088002 536992004 107032001 243224002 
763536836 2052710...

result:

ok 200 lines