dijkstra + 拓扑排序
这道题有负权边，但是卡了spfa，所以我们应该观察题目性质。
负权边一定是单向的，且不构成环，那么我们考虑先将正权边连上。然后dfs一次找到所有正权边构成的联通块，将他们看成点，那么负权边和这些点就构成了一张DAG。
对于DAG，我们可以拓扑排序，在排序的过程中，我们把入度为0的联通块里的所有点松弛一次，如果访问到联通块外的点，就让其入度减1，然后重复在拓扑排序中跑最短路的过程即可得到答案。
输出答案的时候注意一个坑，因为存在负权边，当有的点不能从起点达到的时候，INF可能被负权边松弛。。
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){int X = 0, w = 0; char ch = 0;while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){A ans = 1;for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;return ans;
}const int N = 40005;
int t, r, p, s, cnt, tot, head[N], c[N], d[N], dist[N];
bool vis[N];
struct Edge { int v, next, w; } edge[N<<5];void addEdge(int a, int b, int w){edge[cnt].v = b, edge[cnt].w = w, edge[cnt].next = head[a], head[a] = cnt ++;
}void dfs(int s){vis[s] = true;c[s] = tot;for(int i = head[s]; i != -1; i = edge[i].next){int u = edge[i].v;if(vis[u]) continue;dfs(u);}
}void topSort(){full(dist, INF);queue<int> q;for(int i = 1; i <= tot; i ++){if(!d[i]) q.push(i);}dist[s] = 0;while(!q.empty()){full(vis, false);int cur = q.front(); q.pop();priority_queue< pair<int, int>, vector< pair<int, int> >, greater< pair<int, int> > > pq;for(int i = 1; i <= t; i ++){if(c[i] == cur) pq.push(make_pair(dist[i], i));}while(!pq.empty()){int f = pq.top().second, dis = pq.top().first; pq.pop();if(vis[f]) continue;vis[f] = true;for(int i = head[f]; i != -1; i = edge[i].next){int u = edge[i].v;if(dist[u] > dis + edge[i].w){dist[u] = dis + edge[i].w;if(c[u] == c[f]) pq.push(make_pair(dist[u], u));}if(c[u] != c[f]){if(!--d[c[u]]) q.push(c[u]);}}}}
}int main(){full(head, -1);t = read(), r = read(), p = read(), s = read();for(int i = 0; i < r; i ++){int u = read(), v = read(), c = read();addEdge(u, v, c), addEdge(v, u, c);}for(int i = 1; i <= t; i ++){if(!vis[i]) ++tot, dfs(i);}for(int i = 0; i < p; i ++){int u = read(), v = read(), c = read();addEdge(u, v, c);}for(int cur = 1; cur <= t; cur ++){for(int i = head[cur]; i != -1; i = edge[i].next){int u = edge[i].v;if(c[u] != c[cur]) d[c[u]] ++;}}topSort();for(int i = 1; i <= t; i ++){printf(dist[i] >= 100000000 ? "NO PATH\n" : "%d\n", dist[i]);}return 0;
}