poj 1275 Cashier Employment(差分约束#6)

此题约束条件较多。

 

 #include <cstdio>#include <algorithm>#include <cstdlib> #include <cstring> #define N 30 #define M 100 #define MAXQ 10000 using namespace std; const int inf = (-1u >> 1); int R[24], t[24];int s[N];int d[N], first[N];int u[M], v[M] , w[M], next[M]; bool inq[N];int cnt[N]; int e;int q[MAXQ]; void addE(int x, int y, int c){      next[e] = first[x], first[x] = e;      u[e] = x, v[e] = y;      w[e] = c;       e++;}void init(int sum){          e = 0;         for (int i = 0; i <= 24; i++)              first[i] = -1;         for (int i = 0; i < 16; i++)         {               addE(i, i + 8, R[i + 8]);         }          for (int i = 16; i < 24; i++)         {               addE(i, (i + 8)%24, R[(i + 8)% 24] - sum);         }         for (int i = 0; i < 23; i++)         {             addE(i, i + 1, 0);             addE(i + 1, i, -t[i + 1]);          }         addE(24, 0, 0);         addE(0, 24, -t[0]);          addE(24, 23, sum);}       bool SPFA(int s, int n, int sum){       for (int i = 0; i < n; i++)d[i] =  -inf;       memset(cnt, 0, sizeof(cnt));        d[s] = 0;       int qs, qe;        qs = qe = 0;        for (int i = 0; i < n; i++)inq[i] = false;        q[qe++] = s;        while (qs < qe)       {              int x = q[qs++];               inq[x] = false;              for (int e = first[x]; e != -1; e = next[e])if (d[v[e]] < d[u[e]] + w[e])              {                    d[v[e]] = d[u[e]] + w[e];                     if (!inq[v[e]])                    {                              inq[v[e]] = true;                              q[qe++] = v[e];                              if (++cnt[v[e]] >= n)return false;                     }              }       }       return true;} int main(){      int ca;      scanf("%d", &ca);      int m;  //   FILE* fp = fopen("in.txt", "r");        while (ca--)      {             for (int i = 0; i < 24; i++)             {                 scanf( "%d", R + i);             }              scanf( "%d", &m);             int tmp;             memset(t, 0, sizeof(t));              for (int i = 0; i < m; i++)              {                   scanf( "%d", &tmp);                  t[tmp]++;             }             bool flag = false;              int l = 0, r = m + 1;             int mid , ans = -1;              while (l  + 1 < r)              {                      mid = ((l + r) / 2);                      init(mid);                     flag = SPFA(24, 25, mid);                      if (flag)                          r = mid, ans = mid;                     else l = mid;               }             if ( ans > 0 && ans < m + 1)printf("%d\n", ans);             else puts("No Solution");      }      return 0;}