[CTSC2008]网络管理 [整体二分]

题面

bzoj
luogu

所有事件按时间排序
按值划分下放

把每一个修改
改成一个删除一个插入

对于一个查询
直接查这个段区间有多少合法点
如果查询值大于等于目标值 进入左区间

如果一个查询无解
那么它要求第k大无解
k > 路径长 用lca维护即可

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#define Sqr(x) ((x)*(x))
using namespace std;
const int N = 8e4 + 5;
const int inf = 1e8 + 5;
struct Edge{
    int v, next;	
}edge[N << 1];
int head[N], esize;
inline void addedge(int x, int y){
    edge[++esize] = (Edge){y, head[x]};
    head[x] = esize;	
}
struct Node{
    int t, x, y, w, d;
    bool type;	//0是插入 1是查询
}node[N << 2], p1[N << 2], p2[N << 2];
int n, m, nsize, a[N], ans[N];
int tim, dfn[N], top[N], son[N], fa[N], size[N], rf[N], dep[N];
bool flag[N];

struct BIT{
	int w[N];
	void ins(int x, int d){while(x <= n){w[x] += d; x += x & -x;}}
	int qry(int x){int res = 0; while(x){res += w[x]; x -= x & -x;} return res;}
}bit;

inline void addnode(int x1, int x2, int x3, int x4, int x5, bool x6){
	++nsize, node[nsize].t = x1, node[nsize].x = x2, node[nsize].y = x3, 
	node[nsize].w = x4, node[nsize].d = x5, node[nsize].type = x6;
}

void dfs1(int x, int ff){
	dep[x] = dep[ff] + 1, size[x] = 1, fa[x] = ff;
	for(int i = head[x], vv; ~i; i = edge[i].next){
		vv = edge[i].v; if(vv == ff) continue;
		dfs1(vv, x);
		size[x] += size[vv];
	    if(size[son[x]] < size[vv]) son[x] = vv;	
	}
}

void dfs2(int x, int tp){
    top[x] = tp, dfn[x] = ++tim, rf[tim] = x;
    if(son[x]) dfs2(son[x], tp); else return;
	for(int i = head[x], vv; ~i; i = edge[i].next){
		vv = edge[i].v; if(vv == fa[x] || vv == son[x]) continue;
	    dfs2(vv, vv);
	}
}

int LCA(int x, int y){
    while(top[x] != top[y]){
	    if(dep[top[x]] < dep[top[y]]) swap(x, y);
		x = fa[top[x]];
	}	
	return dep[x] < dep[y] ? x : y;
}

int qry(int x, int y){
    int res = 0;
    while(top[x] != top[y]){
		if(dep[top[x]] < dep[top[y]]) swap(x, y);
		res += bit.qry(dfn[x]) - bit.qry(dfn[top[x]] - 1);
		x = fa[top[x]];
	}	
	if(dep[x] < dep[y]) swap(x, y); res += bit.qry(dfn[x]) - bit.qry(dfn[y] - 1);
	return res;
}

void erfn(int L, int R, int l, int r){
	if(L > R) return;
	if(l == r){
		for(int i = L; i <= R; ++i)
			if(node[i].type && ~ans[node[i].t]) ans[node[i].t] = l;
	    return ;	
	}
	int mid = l + ((r - l) >> 1);
	int t1 = 0, t2 = 0;
	for(int i = L; i <= R; ++i){
	    if(node[i].type) {
			int cnt = qry(node[i].x, node[i].y);
			if(cnt >= node[i].w) p2[++t2] = node[i];
			else node[i].w -= cnt, p1[++t1] = node[i];//第k大!!先减再复制!!
		}
		else {
		    if(node[i].w <= mid) p1[++t1] = node[i];
			else p2[++t2] = node[i], bit.ins(dfn[node[i].x], node[i].d);
		}
	}
	for(int i = L; i <= R; ++i) 
		if(!node[i].type && node[i].w > mid) 
			bit.ins(dfn[node[i].x], -node[i].d);
	for(int i = 1; i <= t1; ++i) node[L + i - 1] = p1[i];
	for(int i = 1; i <= t2; ++i) node[L + t1 + i - 1] = p2[i];	
	erfn(L, L + t1 - 1, l, mid); erfn(L + t1, R, mid + 1, r);
}

int main() {
	memset(head, -1, sizeof(head));
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i){
		scanf("%d", &a[i]);
		addnode(0, i, 0, a[i], 1, 0);
	}
	for(int i = 1, x, y; i < n; ++i){
	    scanf("%d%d", &x, &y);
	    addedge(x, y); addedge(y, x);	
	} dfs1(1, 0); dfs2(1, 1);
	for(int i = 1, x, y, z; i <= m; ++i){
	    scanf("%d%d%d", &x, &y, &z); flag[i] = (bool)x; 
		if(x){ 
			int len = dep[y] + dep[z] - 2 * dep[LCA(y, z)] + 1;
			if(x > len) ans[i] = -1;
			else addnode(i, y, z, x, 0, 1);
		}	
	    else addnode(i, y, 0, a[y], -1, 0), addnode(i, y, 0, z, 1, 0), a[y] = z, ans[i] = -1;
	}
	erfn(1, nsize, 0, inf);
	for(int i = 1; i <= m; ++i)
		if(flag[i]){
			if(ans[i] == -1) printf("invalid request!\n"); 
			else printf("%d\n", ans[i]);
		} 
    //system("PAUSE");
    return 0;
}
posted @ 2019-04-05 11:23  hjmmm  阅读(138)  评论(0编辑  收藏  举报