版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/xiang_6/article/details/81773304
题意:
略
思路:
用拓扑找环的方法,O(n) 把度数为1(跟其相连的点小于2)的点去掉,并标记,那样剩下一些环,然后dfs找每个环的权值和 和 环中结点的个数;将点个数为偶数的环去掉
其实这样做的话可以直接计算 有奇数个点的环的权值;
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <set>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 7;
int n, m;
ll v[maxn];
vector<int> vec[maxn];
ll sum = 0;
int d[maxn];
bool vis[maxn];
void init() {
scanf("%d%d", &n, &m);
sum = 0;
memset(vis, 0, sizeof vis);
memset(d, 0, sizeof d);
for(int i = 1; i <= n; ++i) {
scanf("%lld", &v[i]);
sum += (v[i]);
vec[i].clear();
}
for(int i = 0; i < m; ++i) {
int u, v; scanf("%d%d", &u, &v);
vec[u].push_back(v);
vec[v].push_back(u);
d[u]++; d[v]++;
}
}
ll dfs1(int id) {
d[id]--;
vis[id] = 1;
ll res = v[id];
for(auto i : vec[id]) {
d[i]--;
if(!vis[i] && d[i] < 2) {
res += (dfs1(i));
}
}
return res;
}
ll tmp, cnt;
void dfs2(int id) {
vis[id] = 1;
tmp += v[id], cnt += 1;
for(auto i : vec[id]) {
if(!vis[i]) {
dfs2(i);
}
}
}
void solve() {
for(int i = 1; i <= n; ++i) {
if(vis[i]) continue;
if(d[i] < 2) {
sum -= dfs1(i);
}
}
for(int i = 1; i <= n; ++i) {
if(vis[i]) continue;
tmp = 0, cnt = 0;
dfs2(i);
if(cnt%2 == 0) sum -= tmp;
}
printf("%lld\n", sum);
}
int main() {
int T; scanf("%d", &T);
while(T--) {
init();
solve();
}
return 0;
}