• 洛谷4438 [Hnoi2018]道路 【树形dp】


    题目

    题目太长懒得打

    题解

    HNOI2018惊现普及+/提高?
    由最长路径很短,设(f[i][x][y])表示(i)号点到根有(x)条未修公路,(y)条未修铁路,子树所有乡村不便利值的最小值
    如果(i)为乡村,直接等于公式
    如果(i)不为乡村,枚举修哪边儿子
    (f[i][x][y] = min{f[ls][x + 1][y] + f[rs][x][y],f[ls][x][y] + f[rs][x][y + 1]})

    (ans = f[1][0][0])
    很多(f)用完就丢了,可以滚一下数组省空间

    完了

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #define LL long long int
    #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
    #define REP(i,n) for (int i = 1; i <= (n); i++)
    #define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
    using namespace std;
    const int maxn = 40005,maxm = 100005,INF = 1000000000;
    inline int read(){
    	int out = 0,flag = 1; char c = getchar();
    	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    	return out * flag;
    }
    LL f[110][41][41];
    int n,s[maxn],t[maxn],a[maxn],b[maxn],c[maxn],dep[maxn];
    int st[maxn],top,id[maxn];
    void dfs(int u){
    	if (u >= n){
    		int x = u - n + 1; id[u] = st[top--];
    		for (int i = 0; i <= dep[u]; i++)
    			for (int j = 0; j <= dep[u]; j++)
    				f[id[u]][i][j] = 1ll * c[x] * (a[x] + i) * (b[x] + j);
    		return;
    	}
    	dep[s[u]] = dep[t[u]] = dep[u] + 1;
    	dfs(s[u]); dfs(t[u]);
    	id[u] = st[top--];
    	for (int i = 0; i <= dep[u]; i++)
    		for (int j = 0; j <= dep[u]; j++)
    			f[id[u]][i][j] = min(f[id[s[u]]][i][j] + f[id[t[u]]][i][j + 1],f[id[s[u]]][i + 1][j] + f[id[t[u]]][i][j]);
    	st[++top] = id[s[u]]; st[++top] = id[t[u]];
    }
    int main(){
    	n = read();
    	for (int i = 1; i < n; i++){
    		s[i] = read(); t[i] = read();
    		if (s[i] < 0) s[i] = -s[i] + n - 1;
    		if (t[i] < 0) t[i] = -t[i] + n - 1;
    	}
    	for (int i = 1; i <= n; i++) a[i] = read(),b[i] = read(),c[i] = read();
    	top = 99;
    	REP(i,top) st[i] = i;
    	dfs(1);
    	printf("%lld
    ",f[id[1]][0][0]);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/8974726.html
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