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  • 【HDOJ】1669 Jamie's Contact Groups

    二分+二分图多重匹配。

      1 /* 1669 */
      2 #include <iostream>
      3 #include <string>
      4 #include <map>
      5 #include <queue>
      6 #include <set>
      7 #include <stack>
      8 #include <vector>
      9 #include <deque>
     10 #include <algorithm>
     11 #include <cstdio>
     12 #include <cmath>
     13 #include <ctime>
     14 #include <cstring>
     15 #include <climits>
     16 #include <cctype>
     17 #include <cassert>
     18 #include <functional>
     19 #include <iterator>
     20 #include <iomanip>
     21 using namespace std;
     22 //#pragma comment(linker,"/STACK:102400000,1024000")
     23 
     24 #define sti                set<int>
     25 #define stpii            set<pair<int, int> >
     26 #define mpii            map<int,int>
     27 #define vi                vector<int>
     28 #define pii                pair<int,int>
     29 #define vpii            vector<pair<int,int> >
     30 #define rep(i, a, n)     for (int i=a;i<n;++i)
     31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     32 #define clr                clear
     33 #define pb                 push_back
     34 #define mp                 make_pair
     35 #define fir                first
     36 #define sec                second
     37 #define all(x)             (x).begin(),(x).end()
     38 #define SZ(x)             ((int)(x).size())
     39 #define lson            l, mid, rt<<1
     40 #define rson            mid+1, r, rt<<1|1
     41 
     42 const int maxn = 1005;
     43 const int maxm = 505;
     44 bool M[maxn][maxm];
     45 bool visit[maxm];
     46 int can[maxm][maxm];
     47 int sz[maxm];
     48 int n, gn, bound;
     49 
     50 bool dfs(int u) {
     51     rep(i, 0, gn) {
     52         if (M[u][i] && !visit[i]) {
     53             visit[i] = true;
     54             if (sz[i] < bound) {
     55                 can[i][sz[i]++] = u;
     56                 return true;
     57             }
     58             rep(j, 0, sz[i]) {
     59                 if (dfs(can[i][j])) {
     60                     can[i][j] = u;
     61                     return true;
     62                 }
     63             }
     64         }
     65     }
     66     
     67     return false;
     68 }
     69 
     70 bool judge(int bound_) {
     71     bound = bound_;
     72     memset(sz, 0, sizeof(sz));
     73     rep(i, 0, n) {
     74         memset(visit, false, sizeof(visit));
     75         if (!dfs(i))
     76             return false;
     77     }
     78     
     79     return true;
     80 }
     81 
     82 void solve() {
     83     int l, r, mid;
     84     int ans;
     85     
     86     l = 0;
     87     ans = r = n;
     88     while (l <= r) {
     89         mid = (l + r) >> 1;
     90         if (judge(mid)) {
     91             ans = mid;
     92             r = mid - 1;
     93         } else {
     94             l = mid + 1;
     95         }
     96     }
     97     
     98     printf("%d
    ", ans);
     99 }
    100 
    101 int main() {
    102     ios::sync_with_stdio(false);
    103     #ifndef ONLINE_JUDGE
    104         freopen("data.in", "r", stdin);
    105         freopen("data.out", "w", stdout);
    106     #endif
    107     
    108     char name[20];
    109     int gid;
    110     char ch;
    111     
    112     while (scanf("%d %d", &n, &gn)!=EOF && (n||gn)) {
    113         memset(M, false, sizeof(M));
    114         rep(i, 0, n) {
    115             scanf("%s", name);
    116             while (1) {
    117                 scanf("%d%c", &gid, &ch);
    118                 M[i][gid] = true;
    119                 if (ch == '
    ')
    120                     break;
    121             }
    122         }
    123         solve();
    124     }
    125     
    126     #ifndef ONLINE_JUDGE
    127         printf("time = %d.
    ", (int)clock());
    128     #endif
    129     
    130     return 0;
    131 }

     二分+网络流Dinic也可以解。

      1 /* 1669 */
      2 #include <iostream>
      3 #include <string>
      4 #include <map>
      5 #include <queue>
      6 #include <set>
      7 #include <stack>
      8 #include <vector>
      9 #include <deque>
     10 #include <algorithm>
     11 #include <cstdio>
     12 #include <cmath>
     13 #include <ctime>
     14 #include <cstring>
     15 #include <climits>
     16 #include <cctype>
     17 #include <cassert>
     18 #include <functional>
     19 #include <iterator>
     20 #include <iomanip>
     21 using namespace std;
     22 //#pragma comment(linker,"/STACK:102400000,1024000")
     23 
     24 #define sti                set<int>
     25 #define stpii            set<pair<int, int> >
     26 #define mpii            map<int,int>
     27 #define vi                vector<int>
     28 #define pii                pair<int,int>
     29 #define vpii            vector<pair<int,int> >
     30 #define rep(i, a, n)     for (int i=a;i<n;++i)
     31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     32 #define clr                clear
     33 #define pb                 push_back
     34 #define mp                 make_pair
     35 #define fir                first
     36 #define sec                second
     37 #define all(x)             (x).begin(),(x).end()
     38 #define SZ(x)             ((int)(x).size())
     39 #define lson            l, mid, rt<<1
     40 #define rson            mid+1, r, rt<<1|1
     41 
     42 typedef struct {
     43     int v, c, nxt;
     44 } Edge_t;
     45 
     46 const int INF = 0x3f3f3f3f;
     47 const int maxn = 1005;
     48 const int maxm = 505;
     49 const int maxv = maxn + maxm;
     50 const int maxe = maxn*maxm*2+5;
     51 Edge_t E[maxe];
     52 int F[maxe];
     53 int head[maxv], head_[maxv];
     54 int id[maxm];
     55 int dis[maxv], Q[maxv];
     56 int m;
     57 int n, gn;
     58 int s, t;
     59 
     60 void init() {
     61     memset(head, -1, sizeof(head));
     62     m = 0;
     63     s = 0;
     64     t = n + gn + 1;
     65 }
     66 
     67 void addEdge(int u, int v, int c) {
     68     E[m].v = v;
     69     E[m].c = c;
     70     E[m].nxt = head[u];
     71     head[u] = m++;
     72     
     73     E[m].v = u;
     74     E[m].c = 0;
     75     E[m].nxt = head[v];
     76     head[v] = m++;
     77 }
     78 
     79 bool bfs() {
     80     int l = 0, r = 0;
     81     int u, v, k;
     82     
     83     Q[r++] = s;
     84     memset(dis, 0, sizeof(dis));
     85     dis[s] = 1;
     86     
     87     while (l < r) {
     88         u = Q[l++];
     89         for (k=head[u]; k!=-1; k=E[k].nxt) {
     90             v = E[k].v;
     91             if (!dis[v] && E[k].c>F[k]) {
     92                 dis[v] = dis[u] + 1;
     93                 if (v == t)
     94                     return false;
     95                 Q[r++] = v;
     96             }
     97         }
     98     }
     99     
    100     return true;
    101 }
    102 
    103 int dfs(int u, int val) {
    104     if (u==t || val==0)
    105         return val;
    106     
    107     int ret = 0;
    108     int tmp, v;
    109     
    110     for (int& k=head_[u]; k!=-1; k=E[k].nxt) {
    111         v = E[k].v;
    112         if (dis[v]==dis[u]+1 && E[k].c>F[k] && (tmp=dfs(v, min(val, E[k].c-F[k])))>0) {
    113             F[k] += tmp;
    114             F[k^1] -= tmp;
    115             ret += tmp;
    116             val -= tmp;
    117             if (val == 0)
    118                 break;
    119         }
    120     }
    121     
    122     return ret;
    123 }
    124 
    125 int Dinic() {
    126     int ret = 0, tmp;
    127     
    128     while (1) {
    129         if (bfs())
    130             break;
    131         
    132         memcpy(head_, head, sizeof(head));
    133         while (1) {
    134             tmp = dfs(s, INF);
    135             if (tmp == 0)
    136                 break;
    137             ret += tmp;
    138         }
    139     }
    140     #ifndef ONLINE_JUDGE
    141         printf("Dinic = %d
    ", ret);
    142     #endif
    143     return ret;
    144 }
    145 
    146 bool judge(int bound) {
    147     rep(i, 1, gn+1)
    148         E[id[i]].c = bound;    
    149     memset(F, 0, sizeof(F));
    150     
    151     return Dinic()>=n;
    152 }
    153 
    154 void solve() {
    155     int l = 0, r = n, mid;
    156     int ans = -1;
    157     
    158     while (r >= l) {
    159         mid = (l + r) >> 1;
    160         if (judge(mid)) {
    161             ans = mid;
    162             r = mid - 1;
    163         } else {
    164             l = mid + 1;
    165         }
    166     }
    167     
    168     printf("%d
    ", ans);
    169 }
    170 
    171 int main() {
    172     ios::sync_with_stdio(false);
    173     #ifndef ONLINE_JUDGE
    174         freopen("data.in", "r", stdin);
    175         freopen("data.out", "w", stdout);
    176     #endif
    177     
    178     char name[25];
    179     char ch;
    180     int gid;
    181     
    182     while (scanf("%d %d", &n, &gn)!=EOF && (n||gn)) {
    183         init();
    184         rep(i, 1, n+1) {
    185             scanf("%s", name);
    186             while (1) {
    187                 scanf("%d%c", &gid, &ch);
    188                 addEdge(i, gid+n+1, 1);
    189                 if (ch == '
    ')
    190                     break;
    191             }
    192         }
    193         rep(i, 1, n+1)
    194             addEdge(s, i, 1);
    195         rep(i, 1, gn+1) {
    196             id[i] = m;
    197             addEdge(n+i, t, 1);
    198         }
    199         solve();
    200     }
    201     
    202     
    203     #ifndef ONLINE_JUDGE
    204         printf("time = %d.
    ", (int)clock());
    205     #endif
    206     
    207     return 0;
    208 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/4990790.html
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