UVA 11478 (differential constraint + two points)

Test instructions
Given a direction graph, each edge has a weight, each time you can select a node and an integer, the weight of all the edges ending with V minus D,
Add the weights of all sides with the start of the V plus D
Finally, let the minimum value of all edges be negative and as large as possible.
Code
#include <stdio.h> #include <iostream> #include <algorithm> #include <cstring> #include < Vector> #include <queue>using namespace std;const int N = 1e3;const int inf = 0x3f3f3f3f;int Dist[n],inq[n],cnt[n]    ; struct node{int v,w; Node (int v,int W): V (v), W (w) {};}; vector<node> g[n];void init (int N) {for (int i = 0; i<=n; i++) G[i].clear ();}    BOOL SPFA (int n) {queue<int> q;    Q.push (0);    memset (inq,0,sizeof (INQ));    memset (cnt,0,sizeof (CNT));    memset (dist,inf,sizeof (Dist));    Dist[0] = 0;    Inq[0] = 1;    Cnt[0] = 1;        while (!q.empty ()) {int u = q.front ();        Q.pop ();        Inq[u] = 0;            for (int i = 0; I<g[u].size (); i++) {int v = g[u][i].v, w= g[u][i].w;                if (Dist[v]>dist[u] + W) {Dist[v] = Dist[u] + W;                if (!inq[v]) Q.push (v), inq[v] = 1;            if (++cnt[v]>n) return true; }}} return false;} BOOL Test (int Mid,int N) {for (int i = 1; i<=n; i++) for (int j = 0; J<g[i].size (); j + +) G[I][J].W-=mid;    int ret = SPFA (n);    for (int i = 1, i<=n; i++) for (int j = 0; J<g[i].size (); j + +) G[I][J].W +=mid; return ret;}    int main () {int v,e;        while (~SCANF ("%d%d", &v,&e)) {init (V);        int UB = 0;        int U, V, W;            for (int i = 0; i<e; i++) {scanf ("%d%d%d", &u,&v,&w);            G[u].push_back (Node (v,w));        UB = max (W,UB);        } for (int i = 1; i<=v; i++) G[0].push_back (node (i,0));        int L = 1, R = UB;        if (test (L,V)) puts ("No solution");        else if (!test (ub+1,v)) puts ("Infinite");                else {while (l<r) {int mid = L + (r-l)/2;                if (test (mid,v)) R = mid;            else L = mid + 1;        } printf ("%d\n", L-1); }} return 0;}

UVA 11478 (differential constraint + two points)