标签:blog io ar os sp for div on 问题
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;
}
标签:blog io ar os sp for div on 问题
原文地址:http://www.cnblogs.com/rolight/p/4095922.html