CodeForces 402 E Strictly Positive Matrix

Strictly Positive Matrix

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 ;

}