CodeForces 402 E 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+7; const int N = 2e3 + 10; 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(1){ now = s.top(); s.pop(); belong[now] = scc_cnt; if(now == u) break; } } } void scc(int n){ now_time = scc_cnt = 0; for(int i = 1; i <= n; ++i) if(!belong[i]) dfs(i); } int main(){ memset(head, -1, sizeof(head)); int n, t; scanf("%d", &n); for(int i = 1; i <= n; ++i){ for(int j = 1; j <= n; ++j){ scanf("%d", &t); if(t) add(i, j); } } scc(n); // cout << "?" << endl; if(scc_cnt ^ 1) puts("NO"); else puts("YES"); return 0; }