-网络流-最大流-二分图- [洛谷 PP3254][网络流24题]圆桌问题

题目描述

Link

Solution

建立二分图和超级源汇,将每个源点连向每个单位的编号,容量为这个单位的人数,将每个圆桌的编号连向超级汇点,容量为这个圆桌所能容纳的人的数量,再在每个单位和每个圆桌分别相连,容量为 $1$。

最后跑最大流就可以了。

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#define MAXN 1000002
#define INF 2000000000
int min(int a, int b) {
if (a < b) return a;
else return b;
}
struct Edge {
int v, nx, w;
} e[MAXN];
std::queue <int> q;
int n, m, head[MAXN], ecnt = 1, x, y, z, r, k, dep[MAXN], cur[MAXN], cnt = 1, totp = 0, tot;
void add(int f, int t, int w) {
e[++ecnt] = (Edge) {t, head[f], w};
head[f] = ecnt;
e[++ecnt] = (Edge) {f, head[t], 0};
head[t] = ecnt;
}
bool bfs(int s, int t) {
memset(dep, 0x7f, sizeof(dep));
while (!q.empty()) q.pop();
for (int i = 1; i <= n + m + 2; i++) {
cur[i] = head[i];
}
dep[s] = 0;
q.push(s);
while (!q.empty()) {
int v = q.front();
q.pop();
for (int i = head[v]; i; i = e[i].nx) {
int to = e[i].v;
if (dep[to] > INF && e[i].w) {
dep[to] = dep[v] + 1;
q.push(to);
}
}
}
if (dep[t] < INF) return 1;
else return 0;
}
int dfs(int u, int t, int l) {
if (!l || u == t) return l;
int fl = 0, f;
for (int i = cur[u]; i; i = e[i].nx) {
cur[u] = i;
int to = e[i].v;
if (dep[to] == dep[u] + 1 && (f = dfs(to, t, min(l, e[i].w)))) {
fl += f;
l -= f;
e[i ^ 1].w += f;
e[i].w -= f;
if (!l) break;
}
}
return fl;
}
int Dinic(int s, int t) {
int maxf = 0;
while (bfs(s, t)) {
maxf += dfs(s, t , INF);
}
return maxf;
}
int a[MAXN], b[MAXN];
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
totp += a[i];
}
for (int i = 1; i <= m; i++) {
scanf("%d", &b[i]);
}
r = n + m + 2;
k = n + m + 1;
for (int i = 1; i <= n; i++) {
add(r, i, a[i]);
}
for (int i = n + 1; i <= n + m; i++) {
add(i, k, b[i - n]);
}
for (int i = 1; i <= n; i++) {
for (int j = n + 1; j <= n + m; j++) {
add(i, j, 1);
}
}
tot = Dinic(r, k);
if (tot != totp) {
puts("0");
return 0;
}
puts("1");
for (int i = 1; i <= n; i++) {
for (int j = head[i]; j; j = e[j].nx) {
int to = e[j].v;
if (to != r && e[j].w == 0) {
printf("%d ", e[j].v - n);
}
}
puts("");
}
return 0;
}
文章作者: RiverFun
文章链接: https://stevebraveman.github.io/blog/2019/05/24/84/
版权声明: 本博客所有文章除特别声明外,均采用 CC BY-NC-SA 4.0 许可协议。转载请注明来自 RiverFun

评论