```
fq.push(make_pair(-fans[n], n));//这里加负号
while (fq.size()) {
int x = fq.top().second;
fq.pop();
for (int i = fHead[x]; i; i = fNext[i]) {
int y = fSide[i];
if (fans[y] < fans[x]) {
fans[y] = fans[x];//这句与下句合并
fans[y] = max(fans[y], Price[y]);
fq.push(make_pair(-fans[y], y));//这里加负号
}
}
}
```
这样可以过
by S0CRiA @ 2021-05-29 20:09:28
好在你谷上测了取不取反都能过
by S0CRiA @ 2021-05-29 20:12:30
说一个可能不准确的想法,个人感觉dij就是保证queue单调的spfa(雾
by 犇犇犇犇 @ 2021-05-29 20:51:43