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  • [SWUST1740] 圆桌问题(最大流)

    题目链接:https://www.oj.swust.edu.cn/problem/show/1740

    看了讨论才发现不能多组输入,真坑。

    源点到n个单位,容量是每个单位的人数。

    每个单位到每个桌子有1条边,容量为1。代表每个单位只能有一个人可以在那一张桌子。

    桌子到汇点建边,容量为桌子最多坐的人数。

    越来越熟练啦

      1 #include <bits/stdc++.h>
      2 using namespace std;
      3 
      4 typedef struct Edge {
      5     int u, v, w, next;
      6 }Edge;
      7 
      8 const int inf = 0x7f7f7f7f;
      9 const int maxn = 4444;
     10 
     11 int cnt, dhead[maxn];
     12 int cur[maxn], dd[maxn];
     13 Edge dedge[80001];
     14 int S, T, N;
     15 
     16 void init() {
     17     memset(dhead, -1, sizeof(dhead));
     18     for(int i = 0; i < maxn; i++) dedge[i].next = -1;
     19     S = 0; cnt = 0;
     20 }
     21 
     22 void adde(int u, int v, int w, int c1=0) {
     23     dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 
     24     dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
     25     dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 
     26     dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
     27 }
     28 
     29 bool bfs(int s, int t, int n) {
     30     queue<int> q;
     31     for(int i = 0; i < n; i++) dd[i] = inf;
     32     dd[s] = 0;
     33     q.push(s);
     34     while(!q.empty()) {
     35         int u = q.front(); q.pop();
     36         for(int i = dhead[u]; ~i; i = dedge[i].next) {
     37             if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
     38                 dd[dedge[i].v] = dd[u] + 1;
     39                 if(dedge[i].v == t) return 1;
     40                 q.push(dedge[i].v);
     41             }
     42         }
     43     }
     44     return 0;
     45 }
     46 
     47 int dinic(int s, int t, int n) {
     48     int st[maxn], top;
     49     int u;
     50     int flow = 0;
     51     while(bfs(s, t, n)) {
     52         for(int i = 0; i < n; i++) cur[i] = dhead[i];
     53         u = s; top = 0;
     54         while(cur[s] != -1) {
     55             if(u == t) {
     56                 int tp = inf;
     57                 for(int i = top - 1; i >= 0; i--) {
     58                     tp = min(tp, dedge[st[i]].w);
     59                 }
     60                 flow += tp;
     61                 for(int i = top - 1; i >= 0; i--) {
     62                     dedge[st[i]].w -= tp;
     63                     dedge[st[i] ^ 1].w += tp;
     64                     if(dedge[st[i]].w == 0) top = i;
     65                 }
     66                 u = dedge[st[top]].u;
     67             }
     68             else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
     69                 st[top++] = cur[u];
     70                 u = dedge[cur[u]].v;
     71             }
     72             else {
     73                 while(u != s && cur[u] == -1) {
     74                     u = dedge[st[--top]].u;
     75                 }
     76                 cur[u] = dedge[cur[u]].next;
     77             }
     78         }
     79     }
     80     return flow;
     81 }
     82 
     83 const int maxm = 666;
     84 int n, m;
     85 int b[maxn], c[maxm];
     86 vector<int> path;
     87 
     88 int main() {
     89     // freopen("in", "r", stdin);
     90     scanf("%d%d",&n,&m);
     91     init();
     92     int sum = 0;
     93     for(int i = 1; i <= n; i++) {
     94         scanf("%d", &b[i]);
     95         sum += b[i];
     96     }
     97     for(int i = 1; i <= m; i++) scanf("%d", &c[i]);
     98     S = 0; T = n + m + 1; N = T + 1;
     99     for(int i = 1; i <= n; i++) adde(S, i, b[i]);
    100     for(int i = 1; i <= m; i++) adde(i+n, T, c[i]);
    101     for(int i = 1; i <= n; i++) {
    102         for(int j = 1; j <= m; j++) {
    103             adde(i, j+n, 1);
    104         }
    105     }
    106     int ret = dinic(S, T, N);
    107     if(ret != sum) {
    108         puts("0");
    109         return 0;
    110     }
    111     printf("1
    ");
    112     for(int i = 1; i <= n; i++) {
    113         path.clear();
    114         for(int j = dhead[i]; ~j; j=dedge[j].next) {
    115             if(!dedge[j].w && dedge[j].v) path.push_back(dedge[j].v-n);
    116         }
    117         for(int j = path.size()-1; j >= 0; j--) printf("%d%c", path[j], j==0?'
    ':' ');
    118     }
    119     return 0;
    120 }
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  • 原文地址:https://www.cnblogs.com/kirai/p/6798161.html
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