POJ 3281 Dining (最大流)

Posted 2020-08-14 Recoder

题目链接：http://poj.org/problem?id=3281

引用一下题解：http://www.cnblogs.com/kuangbin/archive/2012/08/21/2649850.html

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 using namespace std;
  5 const int N = 1e3 + 5;
  6 const int M = 1e4 + 5;
  7 const int inf = 0x3f3f3f3f;
  8 struct Edge {
  9     int to, next, cap, flow;
 10 }edge[M];
 11 int tot, head[N], gap[N], dep[N], cur[N];
 12 void init() {
 13     tot = 0;
 14     memset(head, -1, sizeof(head));
 15 }
 16 void addedge(int u, int v, int w, int rw = 0) {
 17     edge[tot].next = head[u]; edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0; head[u] = tot++;
 18     edge[tot].next = head[v]; edge[tot].to = u; edge[tot].cap = rw; edge[tot].flow = 0; head[v] = tot++;
 19 }
 20 int Q[N];
 21 void bfs(int start, int end) {
 22     memset(dep, -1, sizeof(dep));
 23     memset(gap, 0, sizeof(gap));
 24     gap[0] = 1;
 25     int front = 0, rear = 0;
 26     dep[end] = 0;
 27     Q[rear++] = end;
 28     while(front != rear) {
 29         int u = Q[front++];
 30         for(int i = head[u]; ~i; i = edge[i].next) {
 31             int v = edge[i].to;
 32             if(dep[v] != -1)
 33                 continue;
 34             Q[rear++] = v;
 35             dep[v] = dep[u] + 1;
 36             gap[dep[v]]++;
 37         }
 38     }
 39 }
 40 int S[N];
 41 int sap(int start, int end, int N) {
 42     bfs(start, end);
 43     memcpy(cur, head, sizeof(head));
 44     int top = 0;
 45     int u = start;
 46     int ans = 0;
 47     while(dep[start] < N) {
 48         if(u == end) {
 49             int Min = inf;
 50             int inser;
 51             for(int i = 0; i < top; ++i) {
 52                 if(Min > edge[S[i]].cap - edge[S[i]].flow) {
 53                     Min = edge[S[i]].cap - edge[S[i]].flow;
 54                     inser = i;
 55                 }
 56             }
 57             for(int i = 0; i < top; ++i) {
 58                 edge[S[i]].flow += Min;
 59                 edge[S[i] ^ 1].flow -= Min;
 60             }
 61             ans += Min;
 62             top = inser;
 63             u = edge[S[top] ^ 1].to;
 64             continue;
 65         }
 66         bool flag = false;
 67         int v;
 68         for(int i = cur[u]; ~i; i = edge[i].next) {
 69             v = edge[i].to;
 70             if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]) {
 71                 flag = true;
 72                 cur[u] = i;
 73                 break;
 74             }
 75         }
 76         if(flag) {
 77             S[top++] = cur[u];
 78             u = v;
 79             continue;
 80         }
 81         int Min = N;
 82         for(int i = head[u]; ~i; i = edge[i].next) {
 83             if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) {
 84                 Min = dep[edge[i].to];
 85                 cur[u] = i;
 86             }
 87         }
 88         gap[dep[u]]--;
 89         if(!gap[dep[u]])
 90             return ans;
 91         dep[u] = Min + 1;
 92         gap[dep[u]]++;
 93         if(u != start)
 94             u = edge[S[--top] ^ 1].to;
 95     }
 96     return ans;
 97 }
 98 
 99 int main()
100 {
101     int n, f, d;
102     while(scanf("%d %d %d", &n, &f, &d) != EOF) {
103         int m1, m2, num;
104         init();
105         for(int k = 1; k <= n; ++k) {
106             scanf("%d %d", &m1, &m2);
107             for(int i = 1; i <= m1; ++i) {
108                 scanf("%d", &num);
109                 addedge(2 * n + num, k, 1);
110             }
111             for(int i = 1; i <= m2; ++i) {
112                 scanf("%d", &num);
113                 addedge(n + k, 2 * n + f + num, 1);
114             }
115         }
116         for(int i = 1; i <= n; ++i) {
117             addedge(i, n + i, 1);
118         }
119         for(int i = 1; i <= f; ++i) {
120             addedge(0, 2 * n + i, 1);
121         }
122         for(int i = 1; i <= d; ++i) {
123             addedge(2 * n + f + i, 2 * n + f + d + 1, 1);
124         }
125         printf("%d\\n", sap(0, 2 * n + f + d + 1, 2 * n + f + d + 2));
126     }
127     return 0;
128 }