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  • POJ 3281 Dining (最大流)

    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
    ", sap(0, 2 * n + f + d + 1, 2 * n + f + d + 2));
    126     }
    127     return 0;
    128 }
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  • 原文地址:https://www.cnblogs.com/Recoder/p/5965330.html
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