题解：
如果原来的 a[i][j] = 0, 现要 a[i][j] = 1， 那么等于 sum{a[i][k] + a[k][j]} > 1。
如果把a[i][j]视作 i -> j 是否能达到。
那么对于上述的那个方程来说，相当于 i先走到k， k再走到j。 单向边。
所以化简之后，就是询问一幅图是不是只有一个强连通缩点。
代码：
#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod = (int)1e9+;
const int N = 2e3 + ;
const int M = N * N;
int head[N], nt[M], to[M], tot;
void add(int u, int v){
to[tot] = v;
nt[tot] = head[u];
head[u] = tot++;
}
int belong[N], dfn[N], low[N], now_time, scc_cnt;
stack<int> s;
void dfs(int u){
dfn[u] = low[u] = ++now_time;
s.push(u);
for(int i = head[u]; ~i; i = nt[i]){
if(!dfn[to[i]]) dfs(to[i]);
if(!belong[to[i]]) low[u] = min(low[u], low[to[i]]);
}
if(dfn[u] == low[u]){
++scc_cnt;
int now;
while(){
now = s.top(); s.pop();
belong[now] = scc_cnt;
if(now == u) break;
}
}
}
void scc(int n){
now_time = scc_cnt = ;
for(int i = ; i <= n; ++i)
if(!belong[i]) dfs(i);
}
int main(){
memset(head, -, sizeof(head));
int n, t;
scanf("%d", &n);
for(int i = ; i <= n; ++i){
for(int j = ; j <= n; ++j){
scanf("%d", &t);
if(t) add(i, j);
}
}
scc(n);
// cout << "?" << endl;
if(scc_cnt ^ ) puts("NO");
else puts("YES");
return ;
}