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  • Luogu4427 [BJOI2018]求和 (树上差分)

    预处理,树上差分。注意深度减一

    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a))
    #define nR(a,b,c) for(register int a = (b); (a) >= (c); --(a))
    #define Fill(a,b) memset(a, b, sizeof(a))
    #define Max(a,b) ((a) > (b) ? (a) : (b))
    #define Min(a,b) ((a) < (b) ? (a) : (b))
    #define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
    
    //#define ON_DEBUGG
    
    #ifdef ON_DEBUGG
    
    #define D_e_Line printf("
    ----------
    ") 
    #define D_e(x) cout << (#x) << " : " << x << endl
    #define Pause() system("pause")
    #define FileOpen() freopen("in.txt", "r", stdin)
    
    #else
    
    #define D_e_Line ;
    #define D_e(x) ;
    #define Pause() ;
    #define FileOpen() ;
    
    #endif
    using namespace std;
    struct ios{
    	template<typename ATP>inline ios& operator >> (ATP &x){
    		x = 0; int f = 1; char ch;
    		for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
    		while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
    		x *= f;
    		return *this;
    	}
    }io;
    
    const int N = 300007;
    const int mod = 998244353;
    
    #define int long long
    struct Edge{
    	int nxt, pre;
    }e[N << 1];
    int head[N], cntEdge;
    inline void add(int u, int v){
    	e[++cntEdge] = (Edge){ head[u], v}, head[u] = cntEdge;
    }
    
    int maxDep;
    int fa[N], son[N], dep[N], siz[N];
    inline void DFS_First(int u, int father){
    	dep[u] = dep[father] + 1, fa[u] = father, siz[u] = 1;
    	maxDep = Max(maxDep, dep[u]); //re define maxDep
    	for(register int i = head[u]; i; i = e[i].nxt){
    		int v = e[i].pre;
    		if(v == father) continue;
    		DFS_First(v, u);
    		siz[u] += siz[v];
    		if(!son[u] || siz[v] > siz[son[u]]){
    			son[u] = v;
    		}
    	}
    }
    int top[N], rnk[N], dfn[N], dfnIndex;
    inline void DFS_Second(int u, int ancester){
    	top[u] = ancester, dfn[u] = ++dfnIndex, rnk[dfnIndex] = u;
    	if(!son[u]) return;
    	DFS_Second(son[u], ancester);
    	for(register int i = head[u]; i; i = e[i].nxt){
    		int v = e[i].pre;
    		if(v != son[u] && v != fa[u])
    			DFS_Second(v, v);
    	}
    }
    int val[51][N], p[N][51];
    inline int Query(int x, int y, int K){
    	int sum = 0;
    	while(top[x] != top[y]){
    		if(dep[top[x]] < dep[top[y]]) Swap(x, y);
    		sum = (sum + val[K][dfn[x]] - val[K][dfn[top[x]] - 1] + mod) % mod;
    		x = fa[top[x]];
    	}
    	//D_e(sum);
    	if(dep[x] < dep[y]) Swap(x, y);
    	return (sum + val[K][dfn[x]] - val[K][dfn[y] - 1] + mod) % mod;
    }
    
    #undef int 
    int main(){
    #define int long long 
    FileOpen();
    	int n;
    	io >> n;
    	R(i,2,n){
    		int u, v;
    		io >> u >> v;
    		add(u, v);
    		add(v, u);
    	}
    	
    	DFS_First(1, 0);
    	DFS_Second(1, 1);
    
    	--maxDep;
    	R(i,1,n) --dep[i]; // !!!!!!
    	R(i,1,maxDep){
    		p[i][1] = i;
    		R(k,2,50){
    			p[i][k] = p[i][k - 1] * i % mod;
    			//D_e(p[i][k]);
    		}
    	}
    	R(k,1,50){
    		R(i,1,n){
    			val[k][i] = (val[k][i - 1] + p[dep[rnk[i]]][k]) % mod;//, D_e(val[k][i]);
    		}
    	}
    	
    	int Ques;
    	io >> Ques;
        while(Ques--){
        	int x, y, K;
        	io >> x >> y >> K;
        	printf("%lld
    ", Query(x, y, K));
        }
        
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11374866.html
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