zoukankan      html  css  js  c++  java
  • HDU 4467 Graph

    Graph

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2618    Accepted Submission(s): 425

    Problem 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 xth 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 ≤ 105), where n is the number of nodes and m is the number of edges.
    The second line consists of n integers, the ith of which represents the color of the ith 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 ≤ 231 - 1) between u and v (u != v).
    The next line contains only one integer q (1 ≤ q ≤ 105), 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.

    Sample Input

    4 3
    0 0 0 0
    1 2 1
    2 3 2
    3 4 3
    4
    Asksum 0 0
    Change 2
    Asksum 0 0
    Asksum 0 1
    4 3
    0 1 0 0
    1 2 1
    2 3 2
    3 4 3
    4
    Asksum 0 0
    Change 3
    Asksum 0 0
    Asksum 0 1
     
    Sample Output
    Case 1:
    6
    3
    3
    Case 2:
    3
    0
    4
     
    Source
     
    解题:由于数量巨大,我们可以只维护部分结点的信息,其余的结点信息暴力计算就是了。
     
    维护每个节点的sum[u][0] sum[u][1]表示与该结点相邻且是0 的sum 是1的sum
     
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 #define pii pair<int,LL>
     5 const int maxn = 200010;
     6 struct arc {
     7     int u,v;
     8     LL w;
     9     bool operator<(const arc &rhs)const {
    10         return (u < rhs.u || u == rhs.u && v < rhs.v);
    11     }
    12 } e[500010];
    13 LL ans[3],sum[maxn][2];
    14 vector<pii>g[maxn];
    15 int n,m,du[maxn],color[maxn];
    16 bool key[maxn];
    17 void init() {
    18     memset(key,false,sizeof key);
    19     memset(ans,0,sizeof ans);
    20     memset(sum,0,sizeof sum);
    21     memset(du,0,sizeof du);
    22     for(int i = 0; i < maxn; ++i) g[i].clear();
    23 }
    24 void update(int u) {
    25     if(key[u]) {
    26         for(int i = g[u].size()-1; i >= 0; --i) {
    27             sum[g[u][i].first][color[u]] -= g[u][i].second;
    28             sum[g[u][i].first][color[u]^1] += g[u][i].second;
    29         }
    30     } else {
    31         sum[u][0] = sum[u][1] = 0;
    32         for(int i = g[u].size()-1; i >= 0; --i) {
    33             sum[u][color[g[u][i].first]] += g[u][i].second;
    34             if(key[g[u][i].first]) {
    35                 sum[g[u][i].first][color[u]] -= g[u][i].second;
    36                 sum[g[u][i].first][color[u]^1] += g[u][i].second;
    37             }
    38         }
    39     }
    40     ans[color[u]] -= sum[u][0];
    41     ans[color[u] + 1] -= sum[u][1];
    42     ans[color[u]^1] += sum[u][0];
    43     ans[(color[u]^1) + 1] += sum[u][1];
    44     color[u] ^= 1;
    45 }
    46 int main() {
    47     int cs = 1,u,v,q;
    48     while(~scanf("%d%d",&n,&m)) {
    49         printf("Case %d:
    ", cs++);
    50         init();
    51         for(int i = 1; i <= n; ++i)
    52             scanf("%d",color + i);
    53         int bound = round(sqrt(n));
    54         for(int i = 0; i < m; ++i) {
    55             scanf("%d%d%I64d",&e[i].u,&e[i].v,&e[i].w);
    56             if(e[i].u > e[i].v) swap(e[i].u,e[i].v);
    57         }
    58         sort(e,e + m);
    59         int t = 1;
    60         for(int i = 1; i < m; ++i) {
    61             if(e[i].u != e[t-1].u || e[i].v != e[t-1].v) e[t++] = e[i];
    62             else e[t-1].w += e[i].w;
    63         }
    64         m = t;
    65         for(int i = 0; i < m; ++i) {
    66             u = e[i].u;
    67             v = e[i].v;
    68             ans[color[u] + color[v]] += e[i].w;
    69             if(++du[u] > bound) key[u] = true;
    70             if(++du[v] > bound) key[v] = true;
    71         }
    72         for(int i = 0; i < m; ++i) {
    73             u = e[i].u;
    74             v = e[i].v;
    75             sum[u][color[v]] += e[i].w;
    76             sum[v][color[u]] += e[i].w;
    77             if(key[u] && key[v]) {
    78                 g[u].push_back(pii(v,e[i].w));
    79                 g[v].push_back(pii(u,e[i].w));
    80             }
    81             if(!key[u]) g[u].push_back(pii(v,e[i].w));
    82             if(!key[v]) g[v].push_back(pii(u,e[i].w));
    83         }
    84         char cmd[20];
    85         scanf("%d",&q);
    86         while(q--) {
    87             scanf("%s",cmd);
    88             if(cmd[0] == 'A') {
    89                 scanf("%d%d",&u,&v);
    90                 printf("%I64d
    ",ans[u + v]);
    91             } else {
    92                 scanf("%d",&u);
    93                 update(u);
    94             }
    95         }
    96     }
    97     return 0;
    98 }
    View Code
  • 相关阅读:
    pyqt动画的使用
    pyqt 自定义信号
    设计工具- QtDesigner
    样式控制-QSS 样式表
    布局管理之 QStackedLayout (堆 布局)
    布局管理之 QGridLayout (网格布局)
    布局管理之 QFormLayout (表单布局)
    看代码中
    公司同事好坑
    我要多开梦幻手游PC端(梦幻手游PC端多开的简单分析及实现办法)
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4776774.html
Copyright © 2011-2022 走看看