[Codeforces 475B] Strongly Connected City

[题目链接]

         https://codeforces.com/contest/475/problem/B

[算法]

        建图后运行Tarjan算法 ， 判断强连通分量数是否为1

        时间复杂度 : O(NM)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 410

struct edge
{
        int to , nxt;
} e[MAXN * MAXN * 2];

int n , m , timer , cnt , tot , top;
char a[MAXN],b[MAXN];
int head[MAXN],dfn[MAXN * MAXN],low[MAXN * MAXN],s[MAXN * MAXN];
bool instack[MAXN * MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}
inline int id(int x,int y)
{
        return (x - 1) * m + y;
}
inline void tarjan(int u)
{
        low[u] = dfn[u] = ++timer;    
        s[++top] = u;
        instack[u] = true;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (!dfn[v])
                {
                        tarjan(v);
                        low[u] = min(low[u],low[v]);
                } else if (instack[v]) low[u] = min(low[u],dfn[v]);
        }
        if (low[u] == dfn[u])
        {
                cnt++;
                int v;
                do
                {
                      v = s[top--];
                      instack[v] = false;
                } while (u != v);
        }
}

int main()
{
        
        read(n); read(m);
        scanf("%s%s",a + 1,b + 1);
        for (int i = 1; i <= n; i++)
        {
                if (a[i] == '<')
                {
                        for (int j = m; j > 1; j--) addedge(id(i,j),id(i,j - 1));        
                }    else
                {
                        for (int j = 1; j < m; j++) addedge(id(i,j),id(i,j + 1));
                } 
        }
        for (int i = 1; i <= m; i++)
        {
                if (b[i] == 'v')
                {
                        for (int j = 1; j < n; j++) addedge(id(j,i),id(j + 1,i));    
                }    else
                {
                        for (int j = n; j > 1; j--) addedge(id(j,i),id(j - 1,i));
                }
        }
        timer = cnt = 0;
        for (int i = 1; i <= n * m; i++)
        {
                if (!dfn[i])
                        tarjan(i);
        }
        if (cnt == 1) printf("YES
");
        else printf("NO
");
        
        return 0;
    
}