yaoxi-std 的博客

$\text{开}\mathop{\text{卷}}\limits^{ju\check{a}n}\text{有益}$

0%

P2763 试题库问题

P2763 试题库问题

题面

题目链接

解法

这道比较容易建图了。

源点连向每类试题,流量为要选出的题目数量。每类试题向题库中的试题连边,流量$+\infty$。最后每道题向汇点连边,流量为$1$即可。

AC代码

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
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
/**
* @file: P2763.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P2763
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
const int MAXN = 2e3 + 10;
const int MAXM = 1e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Dinic {
struct Edge {
int v, flow;
} edge[MAXM];
int tot = 1, flow = 0;
int head[MAXN], nxt[MAXM], lev[MAXN], cur[MAXN];
void addedge(int u, int v, int flow) {
edge[++tot] = {v, flow};
nxt[tot] = head[u], head[u] = tot;
edge[++tot] = {u, 0};
nxt[tot] = head[v], head[v] = tot;
}
bool bfs(int s, int t) {
memset(lev, -1, sizeof(lev));
queue<int> que;
que.push(s);
lev[s] = 0;
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = head[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (edge[i].flow && lev[v] == -1) {
lev[v] = lev[u] + 1;
que.push(v);
}
}
}
return lev[t] != -1;
}
int augment(int u, int t, int mx) {
if (u == t || mx == 0)
return mx;
int ret = 0;
for (int &i = cur[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (lev[v] != lev[u] + 1)
continue;
int tmp = augment(v, t, min(mx, edge[i].flow));
mx -= tmp, ret += tmp;
edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
if (mx == 0)
break;
}
return ret;
}
int maxflow(int s, int t) {
while (bfs(s, t)) {
memcpy(cur, head, sizeof(cur));
flow += augment(s, t, INF);
}
return flow;
}
};
int k, n, m, s, t, num, a[MAXN];
Dinic maxflow;
signed main() {
read(k), read(n);
s = ++num, t = ++num;
for (int i = 1; i <= k; ++i) {
read(a[i]);
m += a[i];
maxflow.addedge(s, ++num, a[i]);
}
for (int i = 1; i <= n; ++i) {
int p, x;
read(p);
++num;
for (int j = 1; j <= p; ++j) {
read(x);
maxflow.addedge(x + 2, num, INF);
}
maxflow.addedge(num, t, 1);
}
if (maxflow.maxflow(s, t) != m)
return puts("No Solution!"), 0;
for (int i = 1; i <= k; ++i) {
printf("%lld: ", i);
for (int j = maxflow.head[i + 2], c = 0; j; j = maxflow.nxt[j])
if (maxflow.edge[j].flow == INF - 1)
printf("%lld%c", maxflow.edge[j].v - k - 2, " \n"[++c == a[i]]);
}
return 0;
}