P1231 教辅的组成
作者：互联网
传送门：https://www.luogu.org/problemnew/show/P1231
这是一道很不错的网络流入门题，关键在于如何建图。
首先，我们将练习册和源点连一条边权为1的边，然后若书 i 和练习册 j 可以配套，就将连一条从练习册 j 到书 i 边，当然边权还是1。同理，答案和书也是如此，最后再将答案和汇点连一条边权为1的边。
但是这么写还是会有点问题，因为经过一本书的路径可能与很多条，书就被使用了多次，显然不符合题意。这时候我们可以将书 i 拆成书 i1 和 i2，i1 和练习册连边，i2 和答案连边，这样就保证没一本书之用过一次了。
建好图后跑最大流就行了。
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 #include<cmath>
5 #include<cstring>
6 #include<cstdlib>
7 #include<cctype>
8 #include<stack>
9 #include<queue>
10 #include<vector>
11 using namespace std;
12 #define enter printf("\n")
13 #define space printf(" ")
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const db eps = 1e-8;
19 const int maxn = 1e4 + 5;
20 inline ll read()
21 {
22 ll ans = 0;
23 char ch = getchar(), last = ' ';
24 while(!isdigit(ch)) {last = ch; ch = getchar();}
25 while(isdigit(ch))
26 {
27 ans = ans * 10 + ch - '0'; ch = getchar();
28 }
29 if(last == '-') ans = -ans;
30 return ans;
31 }
32 inline void write(ll x)
33 {
34 if(x < 0) x = -x, putchar('-');
35 if(x >= 10) write(x / 10);
36 putchar(x % 10 + '0');
37 }
38
39 int t, n1, n2, n3;
40
41 struct Edge
42 {
43 int from, to, cap, flow;
44 };
45 vector<Edge> edges;
46 vector<int> G[maxn << 2];
47 void addEdge(int from, int to)
48 {
49 edges.push_back((Edge){from, to, 1, 0});
50 edges.push_back((Edge){to, from, 0, 0});
51 int sz = edges.size();
52 G[from].push_back(sz - 2);
53 G[to].push_back(sz - 1);
54 }
55
56 int dis[maxn << 2];
57 bool vis[maxn << 2];
58 bool bfs()
59 {
60 Mem(vis);
61 queue<int> q;
62 q.push(0); vis[0] = 1;
63 dis[0] = 0;
64 while(!q.empty())
65 {
66 int now = q.front(); q.pop();
67 for(int i = 0; i < (int)G[now].size(); ++i)
68 {
69 Edge& e = edges[G[now][i]];
70 if(!vis[e.to] && e.cap > e.flow)
71 {
72 vis[e.to] = 1;
73 dis[e.to] = dis[now] + 1;
74 q.push(e.to);
75 }
76 }
77 }
78 return vis[t];
79 }
80 int cur[maxn << 2];
81 int dfs(int now, int a)
82 {
83 if(now == t || !a) return a;
84 int flow = 0, f;
85 for(int& i = cur[now]; i < (int)G[now].size(); ++i)
86 {
87 Edge& e = edges[G[now][i]];
88 if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
89 {
90 e.flow += f;
91 edges[G[now][i] ^ 1].flow -= f;
92 flow += f; a -= f;
93 if(!a) break;
94 }
95 }
96 return flow;
97 }
98
99 int maxflow()
100 {
101 int flow = 0;
102 while(bfs())
103 {
104 Mem(cur);
105 flow += dfs(0, INF);
106 }
107 return flow;
108 }
109
110 //bool vis2[maxn];
111
112 int main()
113 {
114 n1 = read(); n2 = read(); n3 = read();
115 t = (n1 << 1)+ n2 + n3 + 1; //为了防止编号重复
116 /*按注释掉的写法会WA掉，因为如果一本书多次输入，那么这个书
117 就多次拆点，这一本书就可以使用多次了 */
118 /* int m = read();
119 while(m--)
120 {
121 int x = read(), y = read();
122 addEdge(0, y + (n1 << 1));
123 addEdge(y + (n1 << 1), x);
124 addEdge(x, x + n1);
125 vis2[x] = 1;
126 }
127 m = read();
128 while(m--)
129 {
130 int x = read(), y = read();
131 if(!vis2[x]) addEdge(x, x + n1), vis2[x] = 1;
132 addEdge(x + n1, y + (n1 << 1) + n2);
133 addEdge(y + (n1 << 1) + n2, t);
134 }*/
135 int m = read();
136 while(m--)
137 {
138 int x = read(), y = read();
139 addEdge(y + (n1 << 1), x);
140 }
141 m = read();
142 while(m--)
143 {
144 int x = read(), y = read();
145 addEdge(x + n1, y + (n1 << 1) + n2);
146 }
147 for(int i = 1; i <= n1; ++i) addEdge(i, i + n1);
148 for(int i = 1; i <= n2; ++i) addEdge(0, i + (n1 << 1));
149 for(int i = 1; i <= n3; ++i) addEdge(i + (n1 << 1) + n2, t);
150 write(maxflow()); enter;
151 }
标签：now,ch,int,flow,教辅,P1231,ans,include,组成 来源： https://blog.51cto.com/u_15234622/2830855