题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5044
题意:对于一棵树,可以进行两种操作:
ADD1 U V K:将U到V结点路上的结点的权值都增加K;
ADD2 U V K:将U到V结点路上的边的权值都增加K;
最后输出所有点和所有边的权值。
思路:思路出自http://blog.youkuaiyun.com/hongrock/article/details/39616757
学了下手动扩栈和输入流外挂
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
#include <cstring>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 100010;
const int log = 20;
struct edge
{
int to, id;
edge() {}
edge(int to, int id) : to(to), id(id) {}
};
vector <edge> g[maxn];
long long add[maxn][2], les[maxn], node[maxn], ed[maxn];
int deg[maxn], par[log][maxn], head[maxn], dep[maxn];
int n;
void dfs(int u, int fa)
{
par[0][u] = fa;
deg[u] = 0;
for(int i = 0; i < g[u].size(); i++)
{
edge &e = g[u][i];
int v = e.to;
if(v != fa)
{
deg[u]++;
head[v] = e.id;
dep[v] = dep[u] + 1;
dfs(v, u);
}
}
}
void init()
{
dep[1] = 0;
dfs(1, -1);
for(int k = 0; k + 1 < log; k++)
{
for(int v = 0; v <= n; v++)
{
if(par[k][v] < 0)
par[k + 1][v] = -1;
else
par[k + 1][v] = par[k][par[k][v]];
}
}
}
int lca(int u, int v)
{
if(dep[u] > dep[v]) swap(u, v);
for(int k = 0; k < log; k++)
{
if((dep[v] - dep[u]) >> k & 1)
v = par[k][v];
}
if(u == v) return u;
for(int k = log - 1; k >= 0; k--)
{
if(par[k][u] != par[k][v])
{
u = par[k][u];
v = par[k][v];
}
}
return par[0][u];
}
inline void in(int &x)
{
x = 0;
bool mk = 0;
char c = getchar();
while(c < 48 || c > 57)
{
if(c == '-') mk = 1;
c = getchar();
}
while(c >= 48 && c <= 57)
{
x = x * 10 + c - 48;
c = getchar();
}
if(mk) x = -x;
}
void solve()
{
queue <int> que;
for(int i = 1; i <= n; i++)
if(!deg[i])
que.push(i);
while(!que.empty())
{
int x = que.front();
que.pop();
node[x] = add[x][0];
add[x][0] -= les[x];
int fa = par[0][x];
add[fa][0] += add[x][0];
add[fa][1] += add[x][1];
ed[head[x]] = add[x][1];
if(!(--deg[fa]))
que.push(fa);
}
}
int main()
{
int t;
scanf("%d", &t);
// int size = 256 << 20; // 256MB
// char *p = (char*)malloc(size) + size;
// __asm__("movl %0, %%esp\n" :: "r"(p));
for(int ca = 1; ca <= t; ca++)
{
int q;
printf("Case #%d:\n", ca);
scanf("%d%d", &n, &q);
for(int i = 0; i <= n; i++)
{
g[i].clear();
add[i][0] = add[i][1] = 0;
les[i] = 0;
}
int a, b, c;
char s[10];
for(int i = 1; i < n; i++)
{
in(a), in(b);
g[a].push_back(edge(b, i));
g[b].push_back(edge(a, i));
}
init();
while(q--)
{
scanf(" %s", s);
scanf("%d%d%d", &a, &b, &c);
int x = lca(a, b);
if(s[3] == '1')
{
add[a][0] += c;
add[b][0] += c;
add[x][0] -= c;
les[x] += c;
}
else
{
add[a][1] += c;
add[b][1] += c;
add[x][1] -= c * 2;
}
}
solve();
for(int i = 1; i <= n; i++)
printf("%I64d%c", node[i], i == n ? '\n' : ' ');
for(int i = 1; i < n; i++)
printf("%I64d%c", ed[i], i == n - 1 ? '\n' : ' ');
if(n == 1)
putchar('\n');
}
return 0;
}