【HDOJ】4183 Pahom on Water

就是一个网络流。red结点容量为2，查看最大流量是否大于等于2。对于条件2，把边反向加入建图。条件1，边正向加入建图。

 /* 4183 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef struct {
     int u, v, c, f;
 } Edge_t;

 const int maxn = ;
 Edge_t E[maxn*maxn];
 vi vc[maxn];
 double F[maxn];
 int X[maxn], Y[maxn], R[maxn];
 int pre[maxn], a[maxn], ID[maxn];
 int n, m;

 void addEdge(int u, int v, int c) {
     E[m].u = u;
     E[m].v = v;
     E[m].c = c;
     E[m].f = ;

     E[m+].u = v;
     E[m+].v = u;
     E[m+].c = ;
     E[m+].f = ;

     vc[u].pb(m);
     vc[v].pb(m+);

     m += ;
 }

 bool judge(int i, int j) {
     if (F[i] >= F[j])
         return false;
     if ((R[i]+R[j])*(R[i]+R[j]) > (X[i]-X[j])*(X[i]-X[j])+(Y[i]-Y[j])*(Y[i]-Y[j]))
         return true;
     return false;
 }

 bool bfs(int s, int t) {
     int u, v;
     int sz, k;
     queue<int> Q;

     memset(a, , sizeof(a));
     Q.push(s);
     a[s] = INT_MAX;
     pre[s] = s;

     while (!Q.empty()) {
         u = Q.front();
         Q.pop();
         sz = SZ(vc[u]);
         rep(i, , sz) {
             k = vc[u][i];
             v = E[k].v;
             if (!a[v] && E[k].f<E[k].c) {
                 a[v] = min(a[u], E[k].c-E[k].f);
                 pre[v] = u;
                 ID[v] = k;
                 Q.push(v);
             }
         }
     }

     return a[t]==;
 }

 int EK(int s, int t) {
     int ret = , tmp;
     int u, v, k;

     while () {
         if (bfs(s, t))
             break;

         tmp = a[t];
         for (v=t,u=pre[v]; v!=s; v=u,u=pre[v]) {
             k = ID[v];
             E[k].f += tmp;
             E[k^].f -= tmp;
         }
         ret += tmp;
     }

     return ret;
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int case_n, n_;
     int s, t;
     int ans;

     scanf("%d", &case_n);
     while (case_n--) {
         scanf("%d", &n_);
         n = n_ + n_;
         m = ;

         rep(i, , n)
             vc[i].clr();

         rep(i, , n_) {
             scanf("%lf %d %d %d", &F[i], &X[i], &Y[i], &R[i]);
             if (F[i] == 400.0)
                 s = i;
             else if (F[i] == 789.0)
                 t = i;
         }

         if (judge(s, t)) {
             puts("Game is VALID");
             continue;
         }

         rep(i, , n_) {
             if (i!=s && i!=t)
                 addEdge(i<<, i<<|, );
             rep(j, , i) {
                 if (judge(i, j))
                     addEdge(i<<|, j<<, );
                 else if (judge(j, i))
                     addEdge(j<<|, i<<, );
             }
         }

         addEdge(s<<, s<<|, );

         ans = EK(s<<, t<<);

         if (ans >= )
             puts("Game is VALID");
         else
             puts("Game is NOT VALID");
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }