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。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <string> 5 #include <cstring> 6 #include <algorithm> 7 #include <limits> 8 #include <vector> 9 #include <stack> 10 #include <queue> 11 #include <set> 12 #include <map> 13 #include <bitset> 14 #include <unordered_map> 15 #include <unordered_set> 16 #define lowbit(x) ( x&(-x) ) 17 #define pi 3.141592653589793 18 #define e 2.718281828459045 19 #define INF 0x3f3f3f3f 20 #define LNF 0x3f3f3f3f3f3f3f3f 21 #define HalF (l + r)>>1 22 #define lsn rt<<1 23 #define rsn rt<<1|1 24 #define Lson lsn, l, mid 25 #define Rson rsn, mid+1, r 26 #define QL Lson, ql, qr 27 #define QR Rson, ql, qr 28 #define myself rt, l, r 29 #define pii pair<int, int> 30 #define MP(a, b) make_pair(a, b) 31 using namespace std; 32 typedef unsigned long long ull; 33 typedef unsigned int uit; 34 typedef long long ll; 35 const int maxN = 505; 36 const ll mod = 1023694381; 37 ll qpow(ll a, ll b = mod - 2) 38 { 39 ll ans = 1; 40 while(b) 41 { 42 if(b & 1) ans = ans * a % mod; 43 b >>= 1; 44 a = a * a % mod; 45 } 46 return ans; 47 } 48 int N, M, K; 49 int op[maxN]; 50 ll dp[maxN][maxN], cnt[maxN] = {0}, p[maxN]; 51 int main() 52 { 53 // for(int i = 2; i * i <= mod; i ++) if(mod % i == 0) printf("%d ", i); //prime 54 scanf("%d%d%d", &N, &M, &K); 55 for(int i = 1; i <= N; i ++) scanf("%d", &op[i]); 56 for(int i = 1; i <= N; i ++) for(int j = 1; j <= N; j ++) dp[i][j] = LNF; 57 for(int i = 1; i <= N; i ++) dp[i][i] = 0; 58 for(int i = 1, u, v; i <= M; i ++) 59 { 60 ll w; scanf("%d%d%lld", &u, &v, &w); 61 dp[u][v] = min(dp[u][v], w); 62 dp[v][u] = min(dp[v][u], w); 63 } 64 for(int i = 1; i <= N; i ++) for(int j = 1; j <= N; j ++) for(int k = 1; k <= N; k ++) 65 { dp[j][k] = min(dp[j][k], dp[j][i] + dp[i][k]); } 66 for(int i = 1; i <= N; i ++) 67 { 68 for(int j = 1; j <= N; j ++) 69 { 70 for(int k = j + 1; k <= N; k ++) 71 { 72 if(dp[j][k] == dp[j][i] + dp[i][k]) 73 { 74 cnt[i] ++; 75 } 76 } 77 } 78 } 79 ll All = N * (N - 1) / 2LL; 80 All %= mod; 81 for(int i = 1; i <= N; i ++) 82 { 83 p[i] = (1 - cnt[i] * qpow(All) % mod + mod) % mod; 84 p[i] = qpow(p[i], K); 85 p[i] = (1 - p[i] + mod) % mod; 86 } 87 ll ans = 0; 88 for(int i = 1; i <= N; i ++) 89 { 90 if(op[i]) ans = (ans + 1) % mod; 91 else ans = (ans + p[i]) % mod; 92 } 93 printf("%lld ", ans); 94 return 0; 95 }