题意

题目链接

Sol

数据水的一批，\(O(n^3)\)暴力可过

实际上只要bitset优化一下floyd复杂度就是对的了(\(O(\frac{n^3}{32})\))

还可以缩点之后bitset维护一下连通性，然后对每个联通块之间的分别算，复杂度是\(O(\frac{nm}{32})\)(好像和上面的没区别。。。)

上面代码是floyed

下面的是tarjan

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
} 
int N;
char s[MAXN];
bitset<MAXN> f[MAXN];
int main() {
    N = read();
    for(int i = 1; i <= N; i++) {
        scanf("%s", s + 1);
        for(int j = 1; j <= N; j++) f[i][j] = (s[j] == '1' || (i == j));
    }
    for(int k = 1; k <= N; k++) 
        for(int i = 1; i <= N; i++) 
            if(f[i][k]) f[i] = f[i] | f[k];
    int ans = 0;
    for(int i = 1; i <= N; i++) ans += f[i].count();
    cout << ans;
    return 0;;
}

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 4001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
} 
int N, dfn[MAXN], low[MAXN], tot, vis[MAXN], col[MAXN], siz[MAXN], cnt, inder[MAXN];
stack<int> s;
char str[MAXN];
bitset<MAXN> f[MAXN];
vector<int> v[MAXN], E[MAXN];
void tarjan(int x) {
    dfn[x] = low[x] = ++tot; vis[x] = 1; s.push(x);
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(!dfn[to]) tarjan(to), low[x] = min(low[x], low[to]);
        else if(vis[to]) low[x] = min(low[x], dfn[to]);
    }
    if(low[x] == dfn[x]) {
        int h; cnt++;
        do {
            h = s.top(); s.pop(); vis[h] = 0;
            col[h] = cnt; siz[cnt]++;
        }while(h != x);
    }
}
void Topsort() {
    queue<int> q;
    for(int i = 1; i <= cnt; i++) {
        f[i][i] = 1; 
        if(!inder[i]) q.push(i);
    }
    while(!q.empty()) {
        int p = q.front(); q.pop(); 
        for(int i = 0; i < E[p].size(); i++) {
            int to = E[p][i]; f[to] |= f[p];
            if(!(--inder[to])) q.push(to);
        }
    }
    int ans = 0;
    for(int i = 1; i <= cnt; i++) {
        for(int j = 1; j <= cnt; j++) {
            if(f[i][j]) ans += siz[i] * siz[j];
        }
    }
    printf("%d\n", ans);
}
int main() {
    N = read();
    for(int i = 1; i <= N; i++) {
        scanf("%s", str + 1);
        for(int j = 1; j <= N; j++) if(str[j] == '1' || (i == j)) v[i].push_back(j);
    }
    for(int i = 1; i <= N; i++) 
        if(!dfn[i]) 
            tarjan(i);
    for(int i = 1; i <= N; i++) {
        for(int j = 0; j < v[i].size(); j++) {
            int to = v[i][j];
            if(col[to] != col[i]) E[col[to]].push_back(col[i]), inder[col[i]]++;
        }
    }
    Topsort();
    return 0;;
}