Description

P. T. Tigris is a student currently studying graph theory. One day, when he was studying hard, GS appeared around the corner shyly and came up with a problem: 
Given a graph with n nodes and m undirected weighted edges, every node having one of two colors, namely black (denoted as 0) and white (denoted as 1), you’re to maintain q operations of either kind: 
* Change x: Change the color of x th node. A black node should be changed into white one and vice versa. 
* Asksum A B: Find the sum of weight of those edges whose two end points are in color A and B respectively. A and B can be either 0 or 1. 
P. T. Tigris doesn’t know how to solve this problem, so he turns to you for help.
 

Input

There are several test cases. 
For each test case, the first line contains two integers, n and m (1 ≤ n,m ≤ 10 5), where n is the number of nodes and m is the number of edges. 
The second line consists of n integers, the i th of which represents the color of the i th node: 0 for black and 1 for white. 
The following m lines represent edges. Each line has three integer u, v and w, indicating there is an edge of weight w (1 ≤ w ≤ 2 31 - 1) between u and v (u != v). 
The next line contains only one integer q (1 ≤ q ≤ 10 5), the number of operations. 
Each of the following q lines describes an operation mentioned before. 
Input is terminated by EOF.
 

Output

For each test case, output several lines. 
The first line contains “Case X:”, where X is the test case number (starting from 1). 
And then, for each “Asksum” query, output one line containing the desired answer.
 
题目大意:给n个点m条无向边,每个点有一个颜色(0 or 1),每条边有一个权值。有q个询问,或翻转某个结点的颜色,或询问一个点的颜色为x、另一个点的颜色为y的边的权和。
思路:直接暴力复杂度高达O(qm),完全无法承受。好像没有其他方法,我们只能暴力地优美一点。
用一个sum[]记录答案,询问的时候O(1)输出。
记deg[u]为结点u的度,对于所有deg[v]<deg[u],连一条边v→u。
state[u]记录对于所有deg[v]<deg[u]的v↔u的权值和。
当修改u的时候,对于所有deg[v]<deg[u],利用state直接修改。对于deg[v]≥deg[u],一个一个修改。
对于一个无重边的图来讲,对每个点来说,所有度数大于等于它的点不超过$ O(\sqrt{m}) $个(因为若一个点的度数为$k$,那么度数大于等于它的点最多$x \leq k$个,这些点的度数总和至少是$k^2$,由于$k^2<=2 \times m$,所以$O(x)=O(\sqrt m)$)
那么时间复杂度为$O(q \times \sqrt m )$
这题要去重边(权和相加)。不去重边,2个点10W条边你还得死……
 
代码(3187MS):
 #include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL; const int MAXN = ;
const int MAXE = ; struct Edge {
int from, to, del;
LL w;
void read() {
scanf("%d%d%I64d", &from, &to, &w);
del = ;
if(from > to) swap(from, to);
}
bool operator < (const Edge &rhs) const {
if(from != rhs.from) return from < rhs.from;
return to < rhs.to;
}
bool operator == (const Edge &rhs) const {
return from == rhs.from && to == rhs.to;
}
} edge[MAXE]; LL sum[];
int n, m, q;
int head[MAXN], deg[MAXN], ecnt;
int col[MAXN];
LL state[MAXN][];
int to[MAXE], next[MAXE], dir[MAXE];
LL weight[MAXE]; void init() {
memset(head, , sizeof(head));
memset(sum, , sizeof(sum));
memset(deg, , sizeof(deg));
memset(state, , sizeof(state));
ecnt = ;
} void add_edge(int u, int v, LL w, bool flag) {
to[ecnt] = v; weight[ecnt] = w; dir[ecnt] = flag; next[ecnt] = head[u]; head[u] = ecnt++;
} void change() {
int x, c;
scanf("%d", &x);
c = col[x]; col[x] = !col[x];
sum[c] -= state[x][];
sum[c + ] -= state[x][];
sum[col[x]] += state[x][];
sum[col[x] + ] += state[x][];
for(int p = head[x]; p; p = next[p]) {
int &v = to[p];
if(!dir[p]) {
state[v][c] -= weight[p];
state[v][col[x]] += weight[p];
}
sum[c + col[v]] -= weight[p];
sum[col[x] + col[v]] += weight[p];
}
} char s[]; int main() {
int t = ;
while(scanf("%d%d", &n, &m) != EOF) {
init();
for(int i = ; i <= n; ++i) scanf("%d", &col[i]);
for(int i = ; i <= m; ++i) {
edge[i].read();
sum[col[edge[i].from] + col[edge[i].to]] += edge[i].w;
}
printf("Case %d:\n", ++t);
sort(edge + , edge + m + );
for(int i = ; i < m; ++i) {
if(edge[i] == edge[i + ]) {
edge[i].del = true;
edge[i + ].w += edge[i].w;
} else {
++deg[edge[i].from], ++deg[edge[i].to];
}
}
for(int i = ; i <= m; ++i) {
if(edge[i].del) continue;
if(deg[edge[i].from] < deg[edge[i].to]) {
add_edge(edge[i].from, edge[i].to, edge[i].w, );
state[edge[i].to][col[edge[i].from]] += edge[i].w;
} else if(deg[edge[i].from] > deg[edge[i].to]) {
add_edge(edge[i].to, edge[i].from, edge[i].w, );
state[edge[i].from][col[edge[i].to]] += edge[i].w;
} else {
add_edge(edge[i].from, edge[i].to, edge[i].w, );
add_edge(edge[i].to, edge[i].from, edge[i].w, );
}
}
scanf("%d", &q);
while(q--) {
scanf("%s", s);
if(*s == 'A') {
int x, y;
scanf("%d%d", &x, &y);
printf("%I64d\n", sum[x + y]);
} else change();
}
}
}

代码(3078MS)(稍微改了一下):

 #include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL; const int MAXN = ;
const int MAXE = ; struct Edge {
int from, to, del;
LL w;
void read() {
scanf("%d%d%I64d", &from, &to, &w);
del = ;
if(from > to) swap(from, to);
}
bool operator < (const Edge &rhs) const {
if(from != rhs.from) return from < rhs.from;
return to < rhs.to;
}
bool operator == (const Edge &rhs) const {
return from == rhs.from && to == rhs.to;
}
} edge[MAXE]; LL sum[];
int n, m, q;
int head[MAXN], deg[MAXN], ecnt;
int col[MAXN];
LL state[MAXN][];
int to[MAXE], next[MAXE];
LL weight[MAXE]; void init() {
memset(head, , sizeof(head));
memset(sum, , sizeof(sum));
memset(deg, , sizeof(deg));
memset(state, , sizeof(state));
ecnt = ;
} void add_edge(int u, int v, LL w) {
to[ecnt] = v; weight[ecnt] = w; next[ecnt] = head[u]; head[u] = ecnt++;
} void change() {
int x, c;
scanf("%d", &x);
c = col[x]; col[x] = !col[x];
sum[c] -= state[x][];
sum[c + ] -= state[x][];
sum[col[x]] += state[x][];
sum[col[x] + ] += state[x][];
for(int p = head[x]; p; p = next[p]) {
int &v = to[p];
state[v][c] -= weight[p];
state[v][col[x]] += weight[p];
sum[c + col[v]] -= weight[p];
sum[col[x] + col[v]] += weight[p];
}
} char s[]; int main() {
int t = ;
while(scanf("%d%d", &n, &m) != EOF) {
init();
for(int i = ; i <= n; ++i) scanf("%d", &col[i]);
for(int i = ; i <= m; ++i) {
edge[i].read();
sum[col[edge[i].from] + col[edge[i].to]] += edge[i].w;
}
printf("Case %d:\n", ++t);
sort(edge + , edge + m + );
for(int i = ; i < m; ++i) {
if(edge[i] == edge[i + ]) {
edge[i].del = true;
edge[i + ].w += edge[i].w;
} else {
++deg[edge[i].from], ++deg[edge[i].to];
}
}
for(int i = ; i <= m; ++i) {
if(edge[i].del) continue;
if(deg[edge[i].from] < deg[edge[i].to]) {
add_edge(edge[i].from, edge[i].to, edge[i].w);
state[edge[i].to][col[edge[i].from]] += edge[i].w;
} else {
add_edge(edge[i].to, edge[i].from, edge[i].w);
state[edge[i].from][col[edge[i].to]] += edge[i].w;
}
}
scanf("%d", &q);
while(q--) {
scanf("%s", s);
if(*s == 'A') {
int x, y;
scanf("%d%d", &x, &y);
printf("%I64d\n", sum[x + y]);
} else change();
}
}
}

HDU 4467 Graph(图论+暴力)(2012 Asia Chengdu Regional Contest)的更多相关文章

  1. HDU 4468 Spy(KMP+贪心)(2012 Asia Chengdu Regional Contest)

    Description “Be subtle! Be subtle! And use your spies for every kind of business. ”― Sun Tzu“A spy w ...

  2. 2012 Asia Chengdu Regional Contest

    Browsing History http://acm.hdu.edu.cn/showproblem.php?pid=4464 签到 #include<cstdio> #include&l ...

  3. HDU-4432-Sum of divisors ( 2012 Asia Tianjin Regional Contest )

    Sum of divisors Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  4. HDU 4436 str2int(后缀自动机)(2012 Asia Tianjin Regional Contest)

    Problem Description In this problem, you are given several strings that contain only digits from '0' ...

  5. HDU 4433 locker 2012 Asia Tianjin Regional Contest 减少国家DP

    意甲冠军:给定的长度可达1000数的顺序,图像password像锁.可以上下滑动,同时会0-9周期. 每个操作.最多三个数字连续操作.现在给出的起始序列和靶序列,获得操作的最小数量,从起始序列与靶序列 ...

  6. HDU 4115 Eliminate the Conflict(2-SAT)(2011 Asia ChengDu Regional Contest)

    Problem Description Conflicts are everywhere in the world, from the young to the elderly, from famil ...

  7. HDU 4441 Queue Sequence(优先队列+Treap树)(2012 Asia Tianjin Regional Contest)

    Problem Description There's a queue obeying the first in first out rule. Each time you can either pu ...

  8. HDU 4433 locker(DP)(2012 Asia Tianjin Regional Contest)

    Problem Description A password locker with N digits, each digit can be rotated to 0-9 circularly.You ...

  9. HDU 4431 Mahjong(枚举+模拟)(2012 Asia Tianjin Regional Contest)

    Problem Description Japanese Mahjong is a four-player game. The game needs four people to sit around ...

随机推荐

  1. Zabbix: Database Monitor Installation

    1.     Install ODBC MS SQL Connector On Zabbix Server    https://support.zabbix.com/browse/ZBX-6839 ...

  2. Viewpager实现今日头条顶部导航的功能

    利用简单的Textview 和Viewpager实现滑动.点击换页的效果,效果图如下: 先上布局文件代码: <?xml version="1.0" encoding=&quo ...

  3. FBI树

    题目描述 我们可以把由"0"和"1"组成的字符串分为三类:全"0"串称为B串,全"1"串称为I串,既含"0&q ...

  4. 动态规划(二)HDU1114

    1.题目来源HDU1114 Sample Input 3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4 Sample Output The ...

  5. frame3.5安装出错

    一般是因为禁用了microsoft update,可以在服务里禁用改为手动,之后启动,然后就可以安装

  6. 搭建docker registry (htpasswd 认证)

    1,拉取docker registry 镜像 docker pull registry 2,创建证书存放目录 mkdir -p /home/registry 3,生成CA证书Edit your /et ...

  7. bootstrap Table动态绑定数据并自定义字段显示值

    第一步:我们在官网下载了bootstrap 的文档,并在项目中引入bootstrap table相关js文件,当然,也要记得引入jquery文件 大概如图: 第二步:定义一个table控件 第三步:j ...

  8. 微信小程序横向滚动

    <scroll-view scroll-x="true" style=" white-space: nowrap; display: flex" > ...

  9. MAC和windows开发操作系统环境,解决Maven工程中报 Missing artifact jdk.tools:jdk.tools

    同事使用的是苹果mac,而我们其他人的开发环境是windows jdk1.8 导致同事从git上pull下来的工程,pom文件是直接报错的, windows下的pom文件设置是这样的: <dep ...

  10. 为什么不早点使用 Git...

    教程:https://www.liaoxuefeng.com/wiki/0013739516305929606dd18361248578c67b8067c8c017b000/0013739628770 ...