洛谷P2762 太空飞行计划问题 网络流

洛谷P2762 太空飞行计划问题 网络流 

 

  1 #include<bits/stdc++.h>
  2 #define LL long long
  3 #define GG int
  4 #define For(i, j, k) for(int i=j; i<=k; i++)
  5 #define Dow(i, j, k) for(int i=j; i>=k; i--)
  6 using namespace std;
  7 inline GG read() {
  8     GG x = 0, f = 1;
  9     char ch = getchar();
 10     while(ch<'0'||ch>'9') { if(ch=='-') f = -1; ch = getchar(); }
 11     while(ch>='0'&&ch<='9') { x = x*10+ch-48; ch = getchar(); }
 12     return x * f;
 13 }
 14 inline void write(GG x) {
 15     if(x<0) putchar('-'), x = -x;
 16     if(x>9) write(x/10);
 17     putchar(x%10+48);
 18 }
 19 inline void writeln(GG x) {
 20     write(x); puts("");
 21 }
 22 
 23 const int N = 61, M = 4011, INF = 1e9; 
 24 struct edge{
 25     int to, pre, val; 
 26 }e[M*2];
 27 int head[N*2], cur[N*2], vis[N*2]; 
 28 int nedge; 
 29 
 30 inline void add(int x, int y, int v) {
 31     e[++nedge] = (edge){ y, head[x], v}; 
 32     head[x] = nedge; 
 33 }
 34 
 35 namespace work_1 {
 36     int n, m, S, T; 
 37     int dep[N*2]; 
 38     inline void pre(int nn, int mm, int ss, int tt) {
 39         n = nn; m = mm; S = ss; T = tt; 
 40     }
 41     inline int bfs() {
 42         For(i, 0, n) dep[i] = 0; 
 43         memset(vis, 0, sizeof vis); 
 44         queue<int> Q; 
 45         Q.push(S); 
 46         dep[S] = 1; 
 47         while(!Q.empty()) {
 48             int u = Q.front(); Q.pop(); 
 49             for(int i=head[u]; i!=-1; i=e[i].pre) {
 50                 int v = e[i].to; 
 51                 if(e[i].val>0 && dep[v]==0) {
 52                     dep[v] = dep[u]+1;
 53                     vis[v] = 1;  
 54                     Q.push(v); 
 55                 }
 56             }
 57         }
 58         return dep[T]; 
 59     }
 60     int dfs(int u, int dist) {
 61         if(u == T) return dist; 
 62         for(int& i=cur[u]; i!=-1; i=e[i].pre) {
 63             int v = e[i].to; 
 64             if(e[i].val>0 && dep[v]==dep[u]+1) {
 65                 int flow = dfs(v, min(e[i].val, dist)); 
 66                 if(flow > 0) {
 67                     e[i].val -= flow; 
 68                     e[i^1].val += flow; 
 69                     return flow; 
 70                 }
 71             }
 72         }
 73         return 0; 
 74     }
 75     inline int Dinic() {
 76         int ans = 0; 
 77         while( bfs() ) {
 78             for(int i=0; i<=n; i++) cur[i] = head[i]; 
 79             while(int d = dfs(S, INF)) 
 80                 ans += d;     
 81         }
 82         return ans; 
 83     }
 84 }
 85 
 86 int main() {
 87     int m = read(), n = read(); 
 88     nedge = -1; 
 89     int sum = 0; 
 90     memset(head, -1, sizeof(head)); 
 91     For(i, 1, m) {
 92         int x = read(); 
 93         sum += x; 
 94         add(0, i, x); add(i, 0, 0); 
 95         char ch;  
 96         while((ch=getchar()) != '\n') {
 97             x = ch-'0'; 
 98             while((ch=getchar())&&ch>='0'&&ch<='9') 
 99                 x=x*10+ch-'0'; 
100             add(i, x+m, INF); add(x+m, i, 0); 
101             if(ch=='\n') break; 
102         }
103     }
104     For(i, 1, n) {
105         int x = read(); 
106         add(i+m, m+n+1, x); add(m+n+1, i+m, 0); 
107     }
108     work_1 :: pre(n+m+1, nedge, 0, n+m+1); 
109     int ans = work_1 :: Dinic(); 
110     ans = sum - ans; 
111     For(i, 1, m) if(vis[i]) printf("%d ",i); 
112     puts(""); 
113     For(i, 1, n) if(vis[i+m]) printf("%d ",i); 
114     puts(""); 
115     writeln(ans); 
116 }

 

posted @ 2018-04-08 15:41  third2333  阅读(126)  评论(0)  编辑  收藏  举报