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;
}
原文:http://www.cnblogs.com/Running-Time/p/4806592.html