CCA的期望【牛客练习赛74】【数学】

题目链接

很明显的，我们可以去算每个点变色的概率就是用单位1减去不变色的概率，那么就是这个点变色的概率了，期望就是将这些概率相加就可以了，当然，黑色点本身就是黑色的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define LNF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define pii pair<int, int>
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 505;
const ll mod = 1023694381;
ll qpow(ll a, ll b = mod - 2)
{
    ll ans = 1;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ans;
}
int N, M, K;
int op[maxN];
ll dp[maxN][maxN], cnt[maxN] = {0}, p[maxN];
int main()
{
//    for(int i = 2; i * i <= mod; i ++) if(mod % i == 0) printf("%d
", i);    //prime
    scanf("%d%d%d", &N, &M, &K);
    for(int i = 1; i <= N; i ++) scanf("%d", &op[i]);
    for(int i = 1; i <= N; i ++) for(int j = 1; j <= N; j ++) dp[i][j] = LNF;
    for(int i = 1; i <= N; i ++) dp[i][i] = 0;
    for(int i = 1, u, v; i <= M; i ++)
    {
        ll w; scanf("%d%d%lld", &u, &v, &w);
        dp[u][v] = min(dp[u][v], w);
        dp[v][u] = min(dp[v][u], w);
    }
    for(int i = 1; i <= N; i ++) for(int j = 1; j <= N; j ++) for(int k = 1; k <= N; k ++)
    { dp[j][k] = min(dp[j][k], dp[j][i] + dp[i][k]); }
    for(int i = 1; i <= N; i ++)
    {
        for(int j = 1; j <= N; j ++)
        {
            for(int k = j + 1; k <= N; k ++)
            {
                if(dp[j][k] == dp[j][i] + dp[i][k])
                {
                    cnt[i] ++;
                }
            }
        }
    }
    ll All = N * (N - 1) / 2LL;
    All %= mod;
    for(int i = 1; i <= N; i ++)
    {
        p[i] = (1 - cnt[i] * qpow(All) % mod + mod) % mod;
        p[i] = qpow(p[i], K);
        p[i] = (1 - p[i] + mod) % mod;
    }
    ll ans = 0;
    for(int i = 1; i <= N; i ++)
    {
        if(op[i]) ans = (ans + 1) % mod;
        else ans = (ans + p[i]) % mod;
    }
    printf("%lld
", ans);
    return 0;
}