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题意:有一张无向图,度数小于2的点会被去掉,直到全都大于等于2,问连通块顶点数为奇数的权值和为多少
分析:首先DFS把度数小于2的vis掉,第二次DFS把属于同一个连通块的vis掉,检查是否为奇数个定点,是累加和。用sz[i]表示i点真实还连着的点的个数
代码:
/************************************************ * Author :Running_Time * Created Time :2015/9/13 星期日 15:34:01 * File Name :B.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e4 + 10; const int E = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; vector<int> G[N]; int a[N], sz[N]; bool vis[N]; int n, m, cnt; ll sum; void init(void) { for (int i=1; i<=n; ++i) G[i].clear (); memset (vis, false, sizeof (vis)); } void DFS(int u) { for (int i=0; i<G[u].size (); ++i) { int v = G[u][i]; if (vis[v]) continue; sz[v]--; if (sz[v] <= 1) { vis[v] = true; DFS (v); } } } void DFS2(int u) { sum += a[u]; cnt++; for (int i=0; i<G[u].size (); ++i) { int v = G[u][i]; if (vis[v]) continue; vis[v] = true; DFS2 (v); } } int main(void) { int T; scanf ("%d", &T); while (T--) { scanf ("%d%d", &n, &m); init (); for (int i=1; i<=n; ++i) scanf ("%d", &a[i]); for (int u, v, i=1; i<=m; ++i) { scanf ("%d%d", &u, &v); G[u].push_back (v); G[v].push_back (u); } for (int i=1; i<=n; ++i) sz[i] = G[i].size (); for (int i=1; i<=n; ++i) { if (vis[i]) continue; if (sz[i] <= 1) { vis[i] = true; DFS (i); } } ll ans = 0; for (int i=1; i<=n; ++i) { if (vis[i]) continue; sum = 0; cnt = 0; vis[i] = true; DFS2 (i); if (cnt & 1) ans += sum; } printf ("%I64d\n", ans); } return 0; }
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原文地址:http://www.cnblogs.com/Running-Time/p/4806592.html