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HDU 5044 Tree 树链剖分

树链剖分离线处理所有的增加操作。考虑如果在线性结构上面处理这样的问题,只要把增加区域的起始点+w,结束点的后面一个点-w,最终输出答案的时候只要扫描一遍就好了,现在通过树链剖分把树转化为类似的线性结构,用同样的方法处理即可。

#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <set>#include <bitset>#include <queue>#include <stack>#include <string>#include <iostream>#include <cmath>#include <climits>using namespace std;const int maxn = 1e5 + 100;int tid[maxn], nid[maxn], rank[maxn], rank1[maxn];int son[maxn], siz[maxn], fa[maxn], top[maxn], dep[maxn];int head[maxn], nxt[maxn << 1], v[maxn << 1], u[maxn << 1], ecnt, idcnt;int n, m, addv1[maxn], addv2[maxn], Vval[maxn], Eval[maxn];int eid[maxn], erank[maxn];inline bool scanf_(int &ret) {   char c; int sgn;   if(c=getchar(),c==EOF) return 0; //EOF   while(c!=‘-‘&&(c<‘0‘||c>‘9‘)) c=getchar();   sgn=(c==‘-‘)?-1:1;   ret=(c==‘-‘)?0:(c-‘0‘);   while(c=getchar(),c>=‘0‘&&c<=‘9‘) ret=ret*10+(c-‘0‘);   ret*=sgn;   return 1;} inline void printf_(int x) {   if(x>9) printf_(x/10);   putchar(x%10+‘0‘);}inline void adde(int uu, int vv, int id) {	u[ecnt] = uu; v[ecnt] = vv; nxt[ecnt] = head[uu]; 	eid[ecnt] = id; head[uu] = ecnt++;}inline void init() {	memset(son, -1, sizeof(son));	memset(head, -1, sizeof(head));	memset(addv1, 0, sizeof(addv1));	memset(addv2, 0, sizeof(addv2));	ecnt = idcnt = 0;}void dfs1(int now, int nowfa, int nowdep) {	fa[now] = nowfa, dep[now] = nowdep; siz[now] = 1;	for(int i = head[now]; ~i; i = nxt[i]) if(v[i] != nowfa) {		dfs1(v[i], now, nowdep + 1);		siz[now] += siz[v[i]];		if(son[now] == -1 || siz[v[i]] > siz[son[now]]) {			son[now] = v[i];		}	}}void dfs2(int now, int tp) {	tid[now] = ++idcnt; rank[idcnt] = now;	top[now] = tp;	if(son[now] == -1) return;	dfs2(son[now], tp);	for(int i = head[now]; ~i; i = nxt[i]) 		if(v[i] != fa[now] && v[i] != son[now]) {			dfs2(v[i], v[i]);		}}inline void gao2(int x, int y, int w) {	while(top[x] != top[y]) {		if(dep[top[x]] < dep[top[y]]) swap(x, y);		addv2[tid[top[x]]] += w;		if(top[rank[tid[x] + 1]] == top[x])			addv2[tid[x] + 1] -= w;		x = fa[top[x]];	}	if(dep[x] > dep[y]) swap(x, y);	addv2[tid[x] + 1] += w; 	if(top[rank[tid[y] + 1]] == top[y])		addv2[tid[y] + 1] -= w;}inline void gao1(int x, int y, int w) {	while(top[x] != top[y]) {		if(dep[top[x]] < dep[top[y]]) swap(x, y);		addv1[tid[top[x]]] += w;		if(top[rank[tid[x] + 1]] == top[x])			addv1[tid[x] + 1] -= w;		x = fa[top[x]];	}	if(dep[x] > dep[y]) swap(x, y);	addv1[tid[x]] += w;	if(top[rank[tid[y] + 1]] == top[y])		addv1[tid[y] + 1] -= w;}void calc() {	int val1 = 0, val2 = 0;	for(int i = 1; i <= n; i++) {		val1 += addv1[i]; val2 += addv2[i];		Vval[rank[i]] = val1;		Eval[rank[i]] = val2;		if(top[rank[i + 1]] != top[rank[i]]) {			val1 = val2 = 0;		}	}}int main() {	int __size__ = 256 << 20;	char * __p__ = (char *) malloc(__size__) + __size__;	__asm__("movl %0,%%esp\n"::"r"(__p__));	int T; scanf("%d", &T);	for(int kase = 1; kase <= T; kase++) {		init();		scanf("%d%d", &n, &m);		for(int i = 1; i < n; i++) {			int a, b; scanf("%d%d", &a, &b);			adde(a, b, i); adde(b, a, i);		}		dfs1(1, 1, 1); dfs2(1, 1);		for(int i = 0; i < ecnt; i++) {			if(dep[u[i]] < dep[v[i]]) {				erank[eid[i]] = v[i];			}			else erank[eid[i]] = u[i];		}		char cmd[16]; 		int a, b, y;		getchar();		while(m--) {			getchar(); getchar(); getchar();			cmd[3] = getchar();			scanf_(a); scanf_(b); scanf_(y);			if(cmd[3] == ‘1‘) gao1(a, b, y);			else gao2(a, b, y);		}		calc();		printf("Case #%d:\n", kase);		for(int i = 1; i <= n; i++) {			if(i > 1) putchar(‘ ‘);			printf_(Vval[i]);		}		puts("");		for(int i = 1; i < n; i++) {			if(i > 1) putchar(‘ ‘);			printf_(Eval[erank[i]]);		}		puts("");	}	return 0;}

  

HDU 5044 Tree 树链剖分