题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874

这题有不连通的情况,特别注意。

觉得是存query的姿势不对,用前向星存了一遍,还是T……

 /*
━━━━━┒ギリギリ♂ eye!
┓┏┓┏┓┃キリキリ♂ mind!
┛┗┛┗┛┃\○/
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a))
#define lp p << 1
#define rp p << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onenum(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb; typedef struct Query {
int idx;
int u, v;
Query() {}
Query(int uu, int vv, int ii) : u(uu), v(vv), idx(ii) {}
}Query; typedef struct Edge {
int u, v, w, idx;
int next;
Edge() {}
Edge(int uu, int vv, int ww) : u(uu), v(vv), w(ww) {}
}Edge; typedef struct Ans {
int idx;
int ans;
Ans() {}
Ans(int aa, int ii) :ans(aa), idx(ii) {}
}Ans; const int maxn = ;
const int maxm = ;
int n, m, c, qcnt;
int depth[maxn];
bool vis[maxn];
int pre[maxn];
Ans ans[maxm];
Edge edge[maxn];
Edge q[maxm];
Query qq[maxm];
int qhead[maxm];
int head[maxn];
int ecnt;
int u, v, w; void adde(int u, int v, int w) {
edge[ecnt].u = u;
edge[ecnt].v = v;
edge[ecnt].w = w;
edge[ecnt].next = head[u];
head[u] = ecnt++;
} void addq(int u, int v, int i) {
q[qcnt].u = u;
q[qcnt].v = v;
q[qcnt].idx = i;
q[qcnt].next = qhead[u];
qhead[u] = ecnt++;
} int find(int x) {
return x == pre[x] ? x : pre[x] = find(pre[x]);
} void unite(int x, int y) {
x = find(x);
y = find(y);
if(x != y) pre[y] = x;
} void dfs(int u, int p, int d) {
depth[u] = d;
for(int i = head[u]; ~i; i=edge[i].next) {
int v = edge[i].v;
if(!vis[v] && v != p) {
dfs(v, u, d+edge[i].w);
unite(u, v);
}
}
vis[u] = ;
for(int i = qhead[u]; ~i; i=q[i].next) {
int uu = q[i].u;
int vv = q[i].v;
int idx = q[i].idx;
if((vis[vv] && uu == u) || (vis[uu] && vv == u)) {
ans[idx].ans = depth[vv] + depth[uu];
}
}
} bool cmp(Ans x, Ans y) {
return x.idx < y.idx;
} int ufs[maxn]; int find1(int x) {
return ufs[x] == x ? x : ufs[x] = find1(ufs[x]);
} void unite1(int x, int y) {
x = find1(x);
y = find1(y);
if(x != y) ufs[y] = x;
} inline bool scan_d(int &num) {
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<''||in>'')) in=getchar();
if(in=='-'){ IsN=true;num=;}
else num=in-'';
while(in=getchar(),in>=''&&in<=''){
num*=,num+=in-'';
}
if(IsN) num=-num;
return true;
} int main() {
// FRead();
while(~scanf("%d%d%d",&n,&m,&c)) {
Cls(vis); Cls(depth); Cls(ans); Clr(head, -); Clr(qhead, -);
qcnt = ; Rep(i, n+) pre[i] = i, ufs[i] = i;
Rep(i, m) {
scan_d(u); scan_d(v); scan_d(w);
adde(u, v, w); adde(v, u, w);
unite1(u, v);
}
Rep(i, c) {
scan_d(u); scan_d(v);
addq(u, v, i);
qq[i] = Query(u, v, i);
}
For(i, , n+) if(!vis[i]) dfs(i, -, );
Rep(i, c) {
if(find1(qq[ans[i].idx].u) != find1(qq[ans[i].idx].v)) puts("Not connected");
else printf("%d\n", ans[i].ans);
}
}
RT ;
}

……TARJAN也T了……

 /*
━━━━━┒ギリギリ♂ eye!
┓┏┓┏┓┃キリキリ♂ mind!
┛┗┛┗┛┃\○/
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a))
#define lp p << 1
#define rp p << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onenum(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb; typedef struct Query {
int idx;
int u, v;
Query() {}
Query(int uu, int vv, int ii) : u(uu), v(vv), idx(ii) {}
}Query; typedef struct Edge {
int u, v, w;
int next;
Edge() {}
Edge(int uu, int vv, int ww) : u(uu), v(vv), w(ww) {}
}Edge; typedef struct Ans {
int idx;
int ans;
Ans() {}
Ans(int aa, int ii) :ans(aa), idx(ii) {}
}Ans; const int maxn = ;
const int maxm = ;
int n, m, c, qcnt;
int depth[maxn];
bool vis[maxn];
int pre[maxn];
Query q[maxm];
Ans ans[maxm];
Edge edge[maxn];
int head[maxn];
int ecnt;
int u, v, w; void adde(int u, int v, int w) {
edge[ecnt].u = u;
edge[ecnt].v = v;
edge[ecnt].w = w;
edge[ecnt].next = head[u];
head[u] = ecnt++;
} int find(int x) {
return x == pre[x] ? x : pre[x] = find(pre[x]);
} void unite(int x, int y) {
x = find(x);
y = find(y);
if(x != y) pre[y] = x;
} void dfs(int u, int p, int d) {
depth[u] = d;
for(int i = head[u]; ~i; i=edge[i].next) {
int v = edge[i].v;
if(!vis[v] && v != p) {
dfs(v, u, d+edge[i].w);
unite(u, v);
}
}
vis[u] = ;
Rep(i, qcnt) {
int uu = q[i].u;
int vv = q[i].v;
int idx = q[i].idx;
if((vis[vv] && uu == u) || (vis[uu] && vv == u)) {
ans[idx].idx = idx;
ans[idx].ans = depth[vv] + depth[uu];
}
}
} bool cmp(Ans x, Ans y) {
return x.idx < y.idx;
} int ufs[maxn]; int find1(int x) {
return ufs[x] == x ? x : ufs[x] = find1(ufs[x]);
} void unite1(int x, int y) {
x = find1(x);
y = find1(y);
if(x != y) ufs[y] = x;
} inline bool scan_d(int &num) {
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<''||in>'')) in=getchar();
if(in=='-'){ IsN=true;num=;}
else num=in-'';
while(in=getchar(),in>=''&&in<=''){
num*=,num+=in-'';
}
if(IsN) num=-num;
return true;
} int main() {
// FRead();
while(~scanf("%d%d%d",&n,&m,&c)) {
Cls(vis); Cls(depth); Cls(ans); Clr(head, -);
qcnt = ; Rep(i, n+) pre[i] = i, ufs[i] = i;
Rep(i, m) {
scan_d(u); scan_d(v); scan_d(w);
adde(u, v, w); adde(v, u, w);
unite1(u, v);
}
Rep(i, c) {
scan_d(u); scan_d(v);
q[qcnt++] = Query(u, v, i);
}
For(i, , n+) if(!vis[i]) dfs(i, -, );
Rep(i, c) {
if(find1(q[ans[i].idx].u) != find1(q[ans[i].idx].v)) puts("Not connected");
else printf("%d\n", ans[i].ans);
}
}
RT ;
}

在线胡搞T了,等下写个离线的。

TLE代码:

 /*
━━━━━┒ギリギリ♂ eye!
┓┏┓┏┓┃キリキリ♂ mind!
┛┗┛┗┛┃\○/
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┛┗┛┗┛┃
┓┏┓┏┓┃
┃┃┃┃┃┃
┻┻┻┻┻┻
*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f, sizeof(a))
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onenum(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb; typedef struct Edge {
int u, v, w;
int next;
}Edge;
const int maxn = ;
const int maxm = ;
int n, m, c;
int depth[maxn], fa[maxn];
int pre[maxn];
Edge edge[maxm];
int head[maxn];
int ecnt;
bool vis[maxn]; void adde(int u, int v, int w) {
edge[ecnt].u = u;
edge[ecnt].v = v;
edge[ecnt].w = w;
edge[ecnt].next = head[u];
head[u] = ecnt++;
} int find(int x) {
x == pre[x] ? x : pre[x] = find(pre[x]);
} void unite(int x, int y) {
x = find(x);
y = find(y);
if(x != y) pre[y] = x;
} void dfs(int u, int p, int d) {
vis[u] = ;
depth[u] = d; fa[u] = p;
for(int i = head[u]; ~i; i=edge[i].next) {
int v = edge[i].v;
int w = edge[i].w;
if(!vis[v]) dfs(v, u, d+w);
}
} int lca(int u, int v) {
int s = ;
while(depth[u] > depth[v]) {
s += depth[u];
u = pre[u];
}
while(depth[v] > depth[u]) {
s += depth[v];
v = pre[v];
}
while(u != v) {
s += (depth[u] + depth[v]);
u = pre[u];
v = pre[v];
}
return s;
} inline bool scan_d(int &num) {
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<''||in>'')) in=getchar();
if(in=='-'){ IsN=true;num=;}
else num=in-'';
while(in=getchar(),in>=''&&in<=''){
num*=,num+=in-'';
}
if(IsN) num=-num;
return true;
} int main() {
FRead();
int u, v, w;
while(~scanf("%d%d%d",&n,&m,&c)) {
Clr(head, -); ecnt = ;
Rep(i, n+) pre[i] = i;
Rep(i, m) {
scan_d(u); scan_d(v); scan_d(w);
adde(u, v, w); adde(v, u, w);
unite(u, v);
}
For(i, , n+) {
if(!vis[i]) {
dfs(i, -, );
}
}
W(c) {
scan_d(u); scan_d(v);
if(find(u) != find(v)) puts("Not connected");
else printf("%d\n", lca(u, v));
}
}
RT ;
}

[HDOJ2874]Connections between cities(LCA, 离线tarjan)的更多相关文章

  1. hdu-2874 Connections between cities(lca+tarjan+并查集)

    题目链接: Connections between cities Time Limit: 10000/5000 MS (Java/Others)     Memory Limit: 32768/327 ...

  2. HDU 2874 Connections between cities(LCA Tarjan)

    Connections between cities [题目链接]Connections between cities [题目类型]LCA Tarjan &题意: 输入一个森林,总节点不超过N ...

  3. hdu 2874 Connections between cities [LCA] (lca->rmq)

    Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (J ...

  4. [POJ1330]Nearest Common Ancestors(LCA, 离线tarjan)

    题目链接:http://poj.org/problem?id=1330 题意就是求一组最近公共祖先,昨晚学了离线tarjan,今天来实现一下. 个人感觉tarjan算法是利用了dfs序和节点深度的关系 ...

  5. 近期公共祖先(LCA)——离线Tarjan算法+并查集优化

    一. 离线Tarjan算法 LCA问题(lowest common ancestors):在一个有根树T中.两个节点和 e&sig=3136f1d5fcf75709d9ac882bd8cfe0 ...

  6. LCA离线Tarjan,树上倍增入门题

    离线Tarjian,来个JVxie大佬博客最近公共祖先LCA(Tarjan算法)的思考和算法实现,还有zhouzhendong大佬的LCA算法解析-Tarjan&倍增&RMQ(其实你们 ...

  7. Connections between cities LCA

    Problem Description After World War X, a lot of cities have been seriously damaged, and we need to r ...

  8. HDU 2874 Connections between cities (LCA)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题意是给你n个点,m条边(无向),q个询问.接下来m行,每行两个点一个边权,而且这个图不能有环路 ...

  9. [hdu2874]Connections between cities(LCA+并查集)

    题意:n棵树,求任意两点的最短距离. 解题关键:并查集判断两点是否位于一棵树上,然后求最短距离即可.此题可以直接对全部区间直接进行st表,因为first数组会将连接的两点的区间表示出来. //#pra ...

随机推荐

  1. UpdateData(false) and UpdateData(true)

    数据更新函数: UpdateData(false); 控件的关联变量的值传给控件并改变控件状态(程序--->EXE) UpdateData(true); 控件的状态传给其关联的变量(EXE--- ...

  2. VBS基础篇 - wscript 对象

    一.wscript对象 描述:提供对 Windows 脚本宿主对象模型根对象的访问.详述:WScript 对象是 Windows 脚本宿主对象模型层次结构的根对象.它可在任何脚本文件中使用,不需要特定 ...

  3. ios显示艺术字字体颜色渐变

    UIColor * myColor = [UIColor colorWithPatternImage:[UIImage imageNamed:@"123.jpg"]]; self. ...

  4. 2005: [Noi2010]能量采集 - BZOJ

    Description 栋栋有一块长方形的地,他在地上种了一种能量植物,这种植物可以采集太阳光的能量.在这些植物采集能量后,栋栋再使用一个能量汇集机器把这些植物采集到的能量汇集到一起. 栋栋的植物种得 ...

  5. 2064: 分裂 - BZOJ

    Description 背景: 和久必分,分久必和... 题目描述: 中国历史上上分分和和次数非常多..通读中国历史的WJMZBMR表示毫无压力. 同时经常搞OI的他把这个变成了一个数学模型. 假设中 ...

  6. JavaScript高级---组合模式设计

    一.设计模式 javascript里面给我们提供了很多种设计模式: 工厂.桥.组合.门面.适配器.装饰者.享元.代理.观察者.命令.责任链 在前面我们实现了工厂模式和桥模式 工厂模式 : 核心:为了生 ...

  7. hadoop浅尝 hadoop与hbase交互

    在安装好hbase之后,运行一个与hadoop无关的纯hbase程序成功了. 接着写一个hadoop与hbase进行交互的小程序,这个程序的运行方法依然与前文相同, 即导出jar文件在shell下运行 ...

  8. .NET基础篇——Entity Framework 数据转换层通用类

    在实现基础的三层开发的时候,大家时常会在数据层对每个实体进行CRUD的操作,其中存在相当多的重复代码.为了减少重复代码的出现,通常都会定义一个共用类,实现相似的操作,下面为大家介绍一下Entity F ...

  9. POJ 2253 Frogger (求某两点之间所有路径中最大边的最小值)

    题意:有两只青蛙,a在第一个石头,b在第二个石头,a要到b那里去,每种a到b的路径中都有最大边,求所有这些最大边的最小值.思路:将所有边长存起来,排好序后,二分枚举答案. 时间复杂度比较高,344ms ...

  10. POJ 2771 Guardian of Decency(求最大点独立集)

    该题反过来想:将所有可能发生恋爱关系的男女配对,那么可以带出去的人数应该等于这个二分图的最大独立集 先要做一下预处理,把不符合要求的双方先求出来, company[i][j]表示i.j四个标准都不符合 ...