题目:http://poj.org/problem?id=1986
一开始用vector建树:
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define MAX_VERTEX 40001
#define MAX_EDGE 40000
#define MAX_QUERY 10000
int N, M, Q, dis[MAX_VERTEX];
bool vis[MAX_VERTEX];
//graph
struct Edge{
int x, y, l;
} edge[MAX_EDGE];
vector<int> road[MAX_VERTEX];
//query
struct Query{
int x, y, l;
} query[MAX_QUERY];
vector<int> ques[MAX_VERTEX];
//Union-Find set
int father[MAX_VERTEX];
int Find(int x){
return father[x] != x ? father[x] = Find(father[x]) : x;
}
//functions
void tarjan(int x, int disToRoot)
{
dis[x] = disToRoot;
vis[x] = true;
father[x] = x;
const vector<int>& v = road[x];
for(int i = 0, n = v.size(); i < n; ++i){
const Edge& e = edge[v[i]];
int y = x == e.x ? e.y : e.x;
if(!vis[y]){
tarjan(y, disToRoot + e.l);
father[y] = x;
}
}
for(int i = 0, n = ques[x].size(); i < n; ++i){
Query& q = query[ques[x][i]];
if(q.l) continue;
int y = x == q.x ? q.y : q.x;
if(vis[y]){
q.l = dis[x] + dis[y] - (dis[Find(y)] << 1);
}
}
}
void solve()
{
memset(vis + 1, false, N);
tarjan(1, 0);
for(int i = 0; i < Q; ++i) printf("%d\n", query[i].l);
}
bool input()
{
if(2 != scanf("%d %d", &N, &M)) return false;
char c;
for(int i = 0; i < M; ++i){
scanf("%d %d %d %c", &edge[i].x, &edge[i].y, &edge[i].l, &c);
road[edge[i].x].push_back(i);
road[edge[i].y].push_back(i);
}
scanf("%d", &Q);
for(int i = 0; i < Q; ++i){
scanf("%d %d", &query[i].x, &query[i].y);
ques[query[i].x].push_back(i);
ques[query[i].y].push_back(i);
query[i].l = 0;
}
return true;
}
int main()
{
while(input()) solve();
return 0;
}
再试着用数组邻接表建树:
#include <cstdio>
#include <cstring>
#define MAX_VERTEX 40001
#define MAX_EDGE 40000
#define MAX_QUERY 10000
int N, M, Q, dis[MAX_VERTEX];
bool vis[MAX_VERTEX];
//graph
struct Edge{
int to, len, next;
} edge[MAX_EDGE * 2];
int eIndex;
int pre[MAX_VERTEX];
//query
struct Query{
int to, no, next;
} query[MAX_QUERY * 2];
int ans[MAX_QUERY];
int qIndex;
int que[MAX_VERTEX];
//Union-Find set
int father[MAX_VERTEX];
int Find(int x){
return father[x] != x ? father[x] = Find(father[x]) : x;
}
//functions
void tarjan(int x, int disToRoot)
{
dis[x] = disToRoot;
vis[x] = true;
father[x] = x;
for(int i = pre[x]; i != -1; i = edge[i].next){
int y = edge[i].to, l = edge[i].len;
if(!vis[y]){
tarjan(y, disToRoot + l);
father[y] = x;
}
}
for(int i = que[x]; i != -1; i = query[i].next){
int y = query[i].to, no = query[i].no;
if(vis[y] && ans[no] == -1){
ans[no] = dis[x] + dis[y] - (dis[Find(y)] << 1);
}
}
}
void solve()
{
memset(vis + 1, false, N);
tarjan(1, 0);
for(int i = 0; i < Q; ++i) printf("%d\n", ans[i]);
}
inline void addEdge(int x, int to, int len)
{
edge[eIndex].to = to;
edge[eIndex].len = len;
edge[eIndex].next = pre[x];
pre[x] = eIndex++;
}
inline void addQuery(int x, int to, int no)
{
query[qIndex].to = to;
query[qIndex].no = no;
query[qIndex].next = que[x];
que[x] = qIndex++;
}
bool input()
{
if(2 != scanf("%d %d", &N, &M)) return false;
memset(pre + 1, -1, N << 2);
memset(que + 1, -1, N << 2);
memset(ans, -1, N << 2);
qIndex = eIndex = 0;
char c;
int x, y, l;
for(int i = 0; i < M; ++i){
scanf("%d %d %d %c", &x, &y, &l, &c);
addEdge(x, y, l);
addEdge(y, x, l);
}
scanf("%d", &Q);
for(int i = 0; i < Q; ++i){
scanf("%d %d", &x, &y);
addQuery(x, y, i);
addQuery(y, x, i);
}
return true;
}
int main()
{
while(input()) solve();
return 0;
}
时间提高了5倍左右,空间也节省了1/3,看起来以后建图和建树还是用数组邻接表吧……