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  • [洛谷P4630][APIO2018] Duathlon 铁人两项

    题目大意:给一张无向图,求三元组$(u,v,w)$满足$u->v->w$为简单路径,求个数

    题解:圆方树,缩点后$DP$,因为同一个点双中的点一定地位相同

    卡点:1.$father$数组开小,一不小心就续到了下面的$bool$的$vis$数组中,然后就挂成$98$,因为发现去掉没用的$vis$数组变成$86$,才找到问题

    C++ Code:

    #include <cstdio>
    #include <cstring>
    #define maxn 100010
    #define maxm 200010
    #define N 200010
    
    #define ONLINE_JUDGE
    #define read() R::READ()
    #include <cctype>
    namespace R {
    	int x;
    	#ifdef ONLINE_JUDGE
    	char *ch, op[1 << 28];
    	inline void init() {
    		fread(ch = op, 1, 1 << 28, stdin);
    	}
    	inline int READ() {
    		while (isspace(*ch)) ch++;
    		for (x = *ch & 15, ch++; isdigit(*ch); ch++) x = x * 10 + (*ch & 15);
    		return x;
    	}
    	#else
    	char ch;
    	inline int READ() {
    		ch = getchar();
    		while (isspace(ch)) ch = getchar();
    		for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
    		return x;
    	}
    	#endif
    }
    
    int n, m;
    long long ans;
    inline int min(int a, int b) {return a < b ? a : b;}
    
    namespace Tree {
    	int CNT;
    	int head[N], cnt;
    	struct Edge {
    		int to, nxt;
    	} e[N << 1];
    	void add(int a, int b) {
    		e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
    		e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
    	}
    	
    	long long w[N], f[N], sz[N], pointsz;
    	int fa[N];
    	void dfs(int u) {
    		sz[u] = bool(u <= n);
    		long long tmp = 0;
    		for (int i = head[u]; i; i = e[i].nxt) {
    			int v = e[i].to;
    			if (v != fa[u]) {
    				fa[v] = u;
    				dfs(v);
    				tmp += sz[v] * sz[u];
    				sz[u] += sz[v];
    			}
    		}
    		tmp += sz[u] * (pointsz - sz[u]);
    		ans += w[u] * tmp << 1ll;
    	}
    }
    namespace Graph {
    	int head[maxn], cnt;
    	struct Edge {
    		int to, nxt;
    	} e[maxm << 1];
    	void add(int a, int b) {
    		e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
    		e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
    	}
    	
    	int DFN[maxn], low[maxn], idx, fa[maxn];
    	int S[maxn], top;
    	void Tarjan(int u) {
    		Tree::pointsz++;
    		DFN[u] = low[u] = ++idx;
    		Tree::w[S[++top] = u] = -1;
    		int v;
    		for (int i = head[u]; i; i = e[i].nxt) {
    			v = e[i].to;
    			if (!DFN[v]) {
    				fa[v] = u;
    				Tarjan(v);
    				low[u] = min(low[u], low[v]);
    				if (low[v] >= DFN[u]) {
    					Tree::w[++Tree::CNT] = 1;
    					Tree::add(Tree::CNT, u);
    					do {
    						v = S[top--];
    						Tree::add(Tree::CNT, v);
    						Tree::w[Tree::CNT]++;
    					} while (v != e[i].to);
    				}
    			} else low[u] = min(low[u], DFN[v]);
    		}
    	}
    	inline void tarjan(int n) {
    		for (int i = 1; i <= n; i++) if (!DFN[i]) {
    			Tree::pointsz = 0;
    			Tarjan(i);
    			Tree::dfs(i);
    		}
    	}
    }
    
    int main() {
    	#ifdef ONLINE_JUDGE
    	R::init();
    	#endif
    	Tree::CNT = n = read(), m = read();
    	for (int i = 1; i <= m; i++) Graph::add(read(), read());
    	Graph::tarjan(n);
    	printf("%lld
    ", ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Memory-of-winter/p/9664395.html
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