题解 [ABC155F] Perils in Parallel
作者:互联网
include <bits/stdc++.h>
define N 200005
using namespace std;
int n, m;
int head[N], nex[N << 1], to[N << 1], eid[N << 1], ecnt;
int p[N], s[N];
struct light {
int pos, state;
}lig[N];
bool cmpLight(const light &lhs, const light &rhs) { return lhs.pos < rhs.pos; }
inline int read() {
int x = 0; char ch = getchar();
while(!isdigit(ch)) ch = getchar();
while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return x;
}
inline void addE(int u, int v, int id) {
to[++ecnt] = v, eid[ecnt] = id;
nex[ecnt] = head[u], head[u] = ecnt;
}
int cntOne;
int dif[N], fa[N], fe[N];
bool vis[N];
void dfs(int u) {
if(s[u]) ++dif[u], ++cntOne;
for(int i = head[u], v; i; i = nex[i]) {
v = to[i];
if(fa[v]) continue;
fa[v] = u, fe[v] = eid[i], dfs(v);
dif[u] += dif[v];
}
}
int main() {
n = read(), m = read();
for(int i = 1; i <= n; ++i)
lig[i].pos = read(), lig[i].state = read();
sort(lig + 1, lig + n + 1, cmpLight);
for(int i = 1; i <= n; ++i)
p[i] = lig[i].pos, s[i] = lig[i].state;
for(int i = n + 1; i >= 1; --i)
s[i] = s[i] ^ s[i - 1];
for(int i = 1, l, r; i <= m; ++i) {
l = read(), r = read();
l = lower_bound(p + 1, p + n + 1, l) - p;
r = upper_bound(p + 1, p + n + 1, r) - p - 1;
if(l <= r) {
addE(l, r + 1, i), addE(r + 1, l, i);
}
}
for(int i = 1; i <= n + 1; ++i)
if(!fa[i]) {
fa[i] = i;
dfs(i);
if(cntOne & 1) {
puts("-1");
return 0;
}
}
int cnt = 0;
for(int i = 1; i <= n + 1; ++i)
if(dif[i] & 1)
vis[fe[i]] = true, ++cnt;
printf("%d\n", cnt);
for(int i = 1; i <= m; ++i)
if(vis[i])
printf("%d ", i);
putchar('\n');
return 0;
}
标签:ch,int,题解,pos,ecnt,read,lig,ABC155F,Parallel 来源: https://www.cnblogs.com/Lskkkno1/p/12568984.html