巧妙的解法。

首先直接限定每一個點不走跑$floyd$是$n^4$的，會超時，我們在$floyd$的時候加入分治思想，用$solve(l, r)$表示不考慮$[l, r]$區間的點所能得到的最短路，然後每一層處理一下就好了，直到最後一層就更新到答案中去。

時間復雜度$O(n^3logn)$。

我的代碼在$Windows$下一運行就爆棧。

Code：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 305;
 
const ll inf = 0x3f3f3f3f3f3f3f3f;

int n;
ll ans = 0LL;

struct Matrix {
    ll s[N][N];
    
    inline void init() {
        memset(s, 0x3f, sizeof(s));    
    }
    
    inline ll* operator [] (const int now) {
        return s[now];
    }
    
};

template <typename T>
inline 
 
void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for(; ch > ‘9‘ || ch < ‘0‘; ch = getchar())
        if(ch == ‘-‘) op = -1;
    for(; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

template <typename T>
inline 
 
void chkMin(T &x, T y) {
    if(y < x) x = y;
}

void solve(int l, int r, Matrix now) {
    if(l == r) {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++) 
                if(i != l && j != l) 
                    ans += (now[i][j] == inf) ? -1 : now[i][j]; 
        return;
    }
    
    Matrix tmp = now; 
    int mid = ((l + r) >> 1);
    
    for(int k = mid + 1; k <= r; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                chkMin(tmp[i][j], tmp[i][k] + tmp[k][j]);
    solve(l, mid, tmp);
    
    tmp = now;
    for(int k = l; k <= mid; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                chkMin(tmp[i][j], tmp[i][k] + tmp[k][j]);
    solve(mid + 1, r, tmp);
}

int main() {
    read(n);
    
    Matrix a;
    for(int i = 1; i <= n; i++) 
        for(int j = 1; j <= n; j++) {
            read(a[i][j]);
            if(a[i][j] == -1) a[i][j] = inf;
        }
    
/*    printf("\n");
    for(int i = 1; i <= n; i++, printf("\n"))
        for(int j = 1; j <= n; j++)
            printf("%lld ", a[i][j]);   */
    
    solve(1, n, a);
    
    printf("%lld\n", ans);
    return 0;
}

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