原题传送门
差分约束
考虑如何连边,建立一个超级源点0
对于接下来的关系
然后用
跑最长路
最终
Code:
#include <bits/stdc++.h>
#define maxn 200010
#define LL long long
using namespace std;
struct Edge{
int to, next, len;
}edge[maxn << 1];
int num, head[maxn], n, m, dis[maxn], vis[maxn], tot[maxn];
queue <int> q;
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void addedge(int x, int y, int z){ edge[++num] = (Edge){y, head[x], z}, head[x] = num; }
int main(){
n = read(), m = read();
while (m--){
int opt = read(), x = read(), y = read();
if ((opt == 2 || opt == 4) && x == y) return puts("-1"), 0;
if (opt == 1) addedge(x, y, 0), addedge(y, x, 0);
else if (opt == 2) addedge(x, y, 1);
else if (opt == 3) addedge(y, x, 0);
else if (opt == 4) addedge(y, x, 1);
else addedge(x, y, 0);
}
for (int i = n; i; --i) addedge(0, i, 1);
vis[0] = 1;
q.push(0);
while (!q.empty()){
int u = q.front(); q.pop();
if ((++tot[u]) > n) return puts("-1"), 0;
vis[u] = 0;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (dis[v] < dis[u] + edge[i].len){
dis[v] = dis[u] + edge[i].len;
if (!vis[v]) vis[v] = 1, q.push(v);
}
}
}
LL ans = 0;
for (int i = 1; i <= n; ++i) ans += dis[i];
printf("%lld\n", ans);
return 0;
}