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 
       
         2 using namespace std; 3 typedef long long LL; 4 #define pii pair
        
          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
         
          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 }