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  • 洛谷P4719 【模板】动态dp(ddp LCT)

    题意

    题目链接

    Sol

    动态dp板子题。有些细节还没搞懂,待我研究明白后再补题解。。。

    #include<bits/stdc++.h>
    #define LL long long 
    using namespace std;
    const int MAXN = 1e5 + 10, INF = INT_MAX;
    template<typename A, typename B> inline bool chmax(A &x, B y) {
    	return x < y ? x = y, 1 : 0;
    }
    inline int read() {
    	char c = getchar(); int x = 0, f = 1;
    	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    	return x * f;
    }
    int N, M, a[MAXN], f[MAXN][2];
    struct Ma {
    	LL m[3][3];
    	Ma() {
    		memset(m, -0x3f, sizeof(m));
    	}
    	Ma operator * (const Ma &rhs) const {
    		Ma ans;
    		for(int i = 1; i <= 2; i++)
    			for(int j = 1; j <= 2; j++)
    				for(int k = 1; k <= 2; k++)
    					chmax(ans.m[i][j], m[i][k] + rhs.m[k][j]);
    		return ans;
    	}
    }val[MAXN], sum[MAXN];
    vector<int> v[MAXN];
    int ch[MAXN][2], fa[MAXN];
    #define ls(x) ch[x][0]
    #define rs(x) ch[x][1]
    void update(int k) {
    	sum[k] = val[k];
    	if(rs(k)) sum[k] = sum[k] * sum[rs(k)];
    	if(ls(k)) sum[k] = sum[ls(k)] * sum[k];
    }
    bool ident(int x) {
    	return rs(fa[x]) == x;
    }
    bool isroot(int x) {
    	return ls(fa[x]) != x && rs(fa[x]) != x;
    }
    void connect(int x, int _fa, int tag) {
    	fa[x] = _fa;
    	ch[_fa][tag] = x;
    }
    void rotate(int x) {
    	int y = fa[x], r = fa[y], yson = ident(x), rson = ident(y);
    	int b = ch[x][yson ^ 1];
    	fa[x] = r;
    	if(!isroot(y)) connect(x, r, rson);
    	connect(b, y, yson);
    	connect(y, x, yson ^ 1);
    	update(y); update(x);
    }
    void splay(int x) {
    	for(int y = fa[x]; !isroot(x); rotate(x), y = fa[x])
    		if(!isroot(y))
    			rotate(ident(y) == ident(x) ? y : x);
    }
    void access(int x) {
    	for(int y = 0; x; x = fa[y = x]) {
    		splay(x);
    		if(rs(x)) {
    			val[x].m[1][1] += max(sum[rs(x)].m[1][1], sum[rs(x)].m[2][1]);
    			val[x].m[2][1] += sum[rs(x)].m[1][1];
    		}
    		if(y) {
    			val[x].m[1][1] -= max(sum[y].m[1][1], sum[y].m[2][1]);
    			val[x].m[2][1] -= sum[y].m[1][1];
    		}
    		val[x].m[1][2] = val[x].m[1][1];
    		rs(x) = y;
    		update(x);
    	}
    }
    int Modify(int p, int v) {
    	access(p); splay(p);
    	val[p].m[2][1] -= a[p] - v; 
    	update(p);
    	a[p] = v;
    	splay(1);
    	return max(sum[1].m[1][1], sum[1].m[2][1]);
    }
    void dfs(int x, int _fa) {
    	fa[x] = _fa;
    	f[x][1] = a[x];
    	for(auto &to : v[x]) {
    		if(to == _fa) continue;
    		dfs(to, x);
    		f[x][0] += max(f[to][0], f[to][1]);
    		f[x][1] += f[to][0];
    	}
    	val[x].m[1][1] = f[x][0]; val[x].m[1][2] = f[x][0];
    	val[x].m[2][1] = f[x][1]; val[x].m[2][2] = -INF;
    	update(x);
    //	sum[x] = val[x];
    }
    int main() {
    	//freopen("a.in", "r", stdin);
    	N = read(); M = read();
    	for(int i = 1; i <= N; i++) a[i] = read();
    	for(int i = 1; i < N; i++) {
    		int x = read(), y = read();
    		v[x].push_back(y);
    		v[y].push_back(x);
    	}
    	dfs(1, 0);
    	while(M--) {
    		int x = read(), y = read();
    		printf("%d
    ", Modify(x, y));
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10425674.html
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