题意：
给出一颗边带权的\(n\)个节点的树，问是否存在最短距离为\(k\)的点对。
分析：
最开始做的题是询问最短距离小于等于\(k\)的点对。
我第一反应是下面的方法一。
方法一：
先求出小于等于\(k\)的点对 和 小于\(k\)(也就是小于等于\(k-1\))的点对，然后相减得到等于\(k\)的点对的个数。
方法二：
直接修改之前的统计点对的函数。
容易知道，第一种方法每组查询跑了两遍，所以运行时间是方法二的两倍。
方法一的代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#define MP make_pair
using namespace std;
void read(int& x) {
x = 0;
char c = ' ';
while(c < '0' || c > '9') c = getchar();
while('0' <= c && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
}
typedef pair<int, int> PII;
const int maxn = 10000 + 10;
const int INF = 0x3f3f3f3f;
struct Edge
{
int v, w, nxt;
Edge() {}
Edge(int v, int w, int nxt): v(v), w(w), nxt(nxt) {}
};
int n, k, ans;
vector<int> d, d2;
bool del[maxn];
int ecnt, head[maxn];
Edge edges[maxn * 2];
void AddEdge(int u, int v, int w) {
edges[ecnt] = Edge(v, w, head[u]);
head[u] = ecnt++;
}
int fa[maxn], sz[maxn];
void dfs(int u) {
sz[u] = 1;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(v == fa[u] || del[v]) continue;
fa[v] = u;
dfs(v);
sz[u] += sz[v];
}
}
PII findCenter(int u, int t) {
PII ans(INF, -1);
int m = 0;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(v == fa[u] || del[v]) continue;
ans = min(ans, findCenter(v, t));
m = max(m, sz[v]);
}
m = max(m, t - sz[u]);
ans = min(ans, MP(m, u));
return ans;
}
void getDist(int u, int p, int dist, vector<int>& d) {
d.push_back(dist);
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v, w = edges[i].w;
if(v == p || del[v]) continue;
getDist(v, u, dist + w, d);
}
}
int cntPiars(vector<int>& d) {
int ans = 0;
sort(d.begin(), d.end());
int j = d.size();
for(int i = 0; i < d.size(); i++) {
while(j && d[i] + d[j-1] > k) j--;
ans += j - (j > i ? 1 : 0);
}
return ans;
}
void solve(int u) {
fa[u] = 0;
dfs(u);
int s = findCenter(u, sz[u]).second;
del[s] = true;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(del[v]) continue;
solve(v);
}
d.clear();
d.push_back(0);
for(int i = head[s]; ~i; i = edges[i].nxt) {
int v = edges[i].v, w = edges[i].w;
if(del[v]) continue;
d2.clear();
getDist(v, s, w, d2);
ans -= cntPiars(d2);
d.insert(d.begin(), d2.begin(), d2.end());
}
ans += cntPiars(d);
del[s] = false;
}
int main()
{
while(scanf("%d", &n) == 1 && n) {
ecnt = 0;
memset(head, -1, sizeof(head));
for(int u = 1; u <= n; u++) {
int v, w;
for(read(v); v; read(v)) {
read(w);
AddEdge(u, v, w);
AddEdge(v, u, w);
}
}
for(read(k); k; read(k)) {
ans = 0;
solve(1);
int pre = ans;
ans = 0;
k--;
solve(1);
printf("%s\n", pre - ans > 0 ? "AYE" : "NAY");
}
printf(".\n");
}
return 0;
}
方法二的代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#define MP make_pair
using namespace std;
void read(int& x) {
x = 0;
char c = ' ';
while(c < '0' || c > '9') c = getchar();
while('0' <= c && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
}
typedef pair<int, int> PII;
const int maxn = 10000 + 10;
const int INF = 0x3f3f3f3f;
struct Edge
{
int v, w, nxt;
Edge() {}
Edge(int v, int w, int nxt): v(v), w(w), nxt(nxt) {}
};
int n, k, ans;
vector<int> d, d2;
bool del[maxn];
int ecnt, head[maxn];
Edge edges[maxn * 2];
void AddEdge(int u, int v, int w) {
edges[ecnt] = Edge(v, w, head[u]);
head[u] = ecnt++;
}
int fa[maxn], sz[maxn];
void dfs(int u) {
sz[u] = 1;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(v == fa[u] || del[v]) continue;
fa[v] = u;
dfs(v);
sz[u] += sz[v];
}
}
PII findCenter(int u, int t) {
PII ans(INF, -1);
int m = 0;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(v == fa[u] || del[v]) continue;
ans = min(ans, findCenter(v, t));
m = max(m, sz[v]);
}
m = max(m, t - sz[u]);
ans = min(ans, MP(m, u));
return ans;
}
void getDist(int u, int p, int dist, vector<int>& d) {
d.push_back(dist);
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v, w = edges[i].w;
if(v == p || del[v]) continue;
getDist(v, u, dist + w, d);
}
}
int cntPiars(vector<int>& d) {
int ans = 0;
sort(d.begin(), d.end());
int i = 0, j = d.size() - 1;
while(i < j) {
if(d[i] + d[j] < k) i++;
else if(d[i] + d[j] > k) j--;
else {
if(d[i] == d[j]) {
ans += (j - i + 1) * (j - i) / 2;
break;
} else {
int l = i, r = j;
while(d[i] == d[l]) i++;
while(d[j] == d[r]) j--;
ans += (i - l) * (r - j);
}
}
}
return ans;
}
void solve(int u) {
fa[u] = 0;
dfs(u);
int s = findCenter(u, sz[u]).second;
del[s] = true;
for(int i = head[u]; ~i; i = edges[i].nxt) {
int v = edges[i].v;
if(del[v]) continue;
solve(v);
}
d.clear();
d.push_back(0);
for(int i = head[s]; ~i; i = edges[i].nxt) {
int v = edges[i].v, w = edges[i].w;
if(del[v]) continue;
d2.clear();
getDist(v, s, w, d2);
ans -= cntPiars(d2);
d.insert(d.begin(), d2.begin(), d2.end());
}
ans += cntPiars(d);
del[s] = false;
}
int main()
{
while(scanf("%d", &n) == 1 && n) {
ecnt = 0;
memset(head, -1, sizeof(head));
for(int u = 1; u <= n; u++) {
int v, w;
for(read(v); v; read(v)) {
read(w);
AddEdge(u, v, w);
AddEdge(v, u, w);
}
}
for(read(k); k; read(k)) {
ans = 0;
solve(1);
printf("%s\n", ans ? "AYE" : "NAY");
}
printf(".\n");
}
return 0;
}
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