HDU 4467 Graph

標籤：

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 OutputCase 1:633Case 2:304 

 

Source2012 Asia Chengdu Regional Contest 解題：由於數量巨大，我們可以只維護部分結點的資訊，其餘的結點資訊暴力計算就是了。 維護每個節點的sum[u][0] sum[u][1]表示與該結點相鄰且是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:\n", 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\n",ans[u + v]);91             } else {92                 scanf("%d",&u);93                 update(u);94             }95         }96     }97     return 0;98 }

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HDU 4467 Graph