Farm of P1993 small K-differential constraints

After the inequality is seen, the model of moving the sleeve
The approximate inequality model is this:\ (x_v <= X_u + w_{(U, v)}\)
\ (X_v-x_u <= w_{(U, v)}\)
From the meiosis to the connected edge
With the shortest-circuited triangular inequalities.
So the solution also revolves around inequalities
It can be seen that the maximum value of \ (x_v\) is a shortest form
If the shortest path does not exist (negative ring), the\ (x_v\) value does not exist
Use DFS_SPFA to judge the shortest way, notice the exit recursion layer when let vis[x] = 0
Also note that the shortest path when initializing the D array is the INF, although the negative ring to initialize to 0 seems to be no problem (also may be the data water), but the initialization of the complexity of the impact is not small ...

#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include < queue>using namespace std; #define DEBUG (x) cerr << #x << "=" << x << endl;const int MAXN = 1000    0 + 10;int n,m,vis[maxn],cnt[maxn],last[maxn],edge_tot,d[maxn];struct edge{int u, V, W, to; Edge () {} edge (int u, int v, int w, int to): U (U), V (v), W (W), to (to) {}}E[MAXN * 2];inline void Add (int u, int v, int    W) {E[++edge_tot] = Edge (U, V, W, Last[u]); Last[u] = Edge_tot;    }bool flg;queue<int> q;void SPFA (int x) {vis[x] = 1;    if (FLG) return;        for (int i=last[x]; i; i=e[i].to) {int v = e[i].v, w = E[I].W;                if (D[v] > D[x] + W) {if (Vis[v]) {FLG = true;//is accessed again in the recursive layer, and the path is shorter, is it possible to walk this way more than once?            Return            } D[v] = d[x] + W;        SPFA (v); }} Vis[x] = 0;}    int main () {scanf ("%d%d", &n, &m); for (int i=1; i<=m; i++) {iNT CMD, A, B, C;        scanf ("%d", &cmd);            if (cmd = = 1) {scanf ("%d%d%d", &a, &b, &c);        Add (A, B,-C);            } else if (cmd = = 2) {scanf ("%d%d%d", &a, &b, &c);        Add (b, A, c);            } else {scanf ("%d%d", &a, &b);            Add (A, b, 0);        Add (b, a, 0);    }} memset (d, 0x3f, sizeof (d));        for (int i=1; i<=n; i++) {d[i] = 0;        SPFA (i);    if (FLG) break;    } if (FLG) printf ("No");    else printf ("Yes"); return 0;}

In addition, the BFS award negative ring (the upper bound of Complexity is n*m):

bool SPFA(int s) {    dist[s] = 0;    q.push(s);    vis[s] = 1;    cnt[s] = 0;    while(!q.empty()) {        int x = q.front();        q.pop();        vis[x] = 0;        for(int i=last[x]; i; i=e[i].to) {            int v = e[i].v;            int w = e[i].w;            if(dist[v] > dist[x] + w) {                dist[v] = dist[x] + w;                cnt[v] = cnt[x] + 1;                if(cnt[v] >= n) {                    return 1;                  }                if(!vis[v]) q.push(v), vis[v] = 1;            }        }    }    return 0;}

Farm of P1993 small K-differential constraints