HDU 5438 Ponds dfs模拟

2015 ACM/ICPC Asia Regional Changchun Online

题意：n个池塘，删掉度数小于2的池塘，输出池塘数为奇数的连通块的池塘容量之和.

思路：两个dfs模拟就行了

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <fstream>
  4 #include <algorithm>
  5 #include <cmath>
  6 #include <deque>
  7 #include <vector>
  8 #include <queue>
  9 #include <string>
 10 #include <cstring>
 11 #include <map>
 12 #include <stack>
 13 #include <set>
 14 #define LL long long
 15 #define eps 1e-8
 16 #define INF 0x3f3f3f3f
 17 #define MAXN 10005
 18 using namespace std;
 19 vector<int> G[MAXN];
 20 int  father[MAXN], cnt[MAXN], du[MAXN];
 21 LL v[MAXN];
 22 bool rm[MAXN], vis[MAXN];
 23 struct Node{
 24     int pos, du;
 25     Node(int pos = 0, int du = 0) :pos(pos), du(du){};
 26 };
 27 bool compare(Node x, Node y){
 28     return x.du < y.du;
 29 }
 30 int find(int x){
 31     if (father[x] == x) return x;
 32     father[x] = find(father[x]);
 33     return father[x];
 34 }
 35 void dfs(int x, int p){
 36     vis[x] = true;
 37     father[x] = p;
 38     for (int i = 0; i < G[x].size(); i++){
 39         int y = G[x][i];
 40         if (rm[y] || vis[y]) continue;
 41         dfs(y, p);
 42     }
 43 }
 44 void dfs_rm(int x){
 45     rm[x] = true;
 46     for (int i = 0; i < G[x].size(); i++){
 47         du[G[x][i]]--;
 48         if (du[G[x][i]] < 2 && !rm[G[x][i]]) dfs_rm(G[x][i]);
 49     }
 50 }
 51 Node node[MAXN];
 52 int main()
 53 {
 54     //freopen("in.txt", "r", stdin);
 55     int T;
 56     int m, n;
 57     scanf("%d", &T);
 58     while (T--){
 59         scanf("%d%d", &n, &m);
 60         memset(du, 0, sizeof(du));
 61         for (int i = 1; i <= n; i++){
 62             scanf("%I64d", &v[i]);
 63             G[i].clear();
 64         }
 65         int x, y;
 66         for (int i = 1; i <= m; i++){
 67             scanf("%d%d", &x, &y);
 68             du[x]++;
 69             du[y]++;
 70             G[x].push_back(y);
 71             G[y].push_back(x);
 72         }
 73         memset(rm, 0, sizeof(rm));
 74         //sort(node + 1, node + n + 1, compare);
 75         for (int i = 1; i <= n; i++){
 76             //int x = node[i].pos;
 77             if (du[i] > 1 || rm[i]) continue;
 78             dfs_rm(i);
 79         }
 80         int p = 1;
 81         memset(father, 0, sizeof(father));
 82         memset(vis, 0, sizeof(vis));
 83         for (int i = 1; i <= n; i++){
 84             if (rm[i] || vis[i]) continue;
 85             dfs(i, p);
 86             p++;
 87         }
 88         memset(cnt, 0, sizeof(cnt));
 89         for (int i = 1; i <= n; i++){
 90             if (rm[i]) continue;
 91             //int x = find(i);
 92             cnt[father[i]]++;
 93         }
 94         LL ans = 0;
 95         for (int i = 1; i <= n; i++){
 96             if (cnt[father[i]] & 1){
 97                 ans += v[i];
 98             }
 99         }
100         printf("%I64d\n", ans);
101     }
102 }

 

posted on 2015-09-13 14:29  张济  阅读(179)  评论(0)  编辑  收藏  举报

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