匹配:模板

匹配:模板

UOJ78 二分图最大匹配(DFS - KM)

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
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

const int N = 550;

int match[N], g[N][N], vis[N], link[N];
int n, m, e, tag, ans = 0;

bool dfs(int u) {

for(int v=1; v<=m; ++v) {
if(!g[u][v] or vis[v] == tag) continue;
vis[v] = tag;
if(!match[v] or dfs(match[v])) {
match[v] = u;
return true;
}
}
return false;
}
int main() {

scanf("%d%d%d", &n, &m, &e);
for(int i=1, u, v; i<=e; ++i) {
scanf("%d%d", &u, &v);
g[u][v] = true;
}

for(int i=1; i<=n; ++i) {
tag = i;
if(dfs(i)) ++ans;
}

printf("%d\n", ans);
for(int i=1; i<=m; ++i) link[match[i]] = i;

for(int i=1; i<=n; ++i) printf("%d ", link[i]);

return 0;
}

UOJ79 一般图最大匹配(带花树)

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
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
using namespace std;

const int N = 550;
int fa[N], match[N], pre[N], vis[N];
int tag = 0, n, m, ans = 0;
int flag[N];
bool g[N][N];
queue<int> q;

void clear(queue<int> &q) {
queue<int> empty;
swap(q, empty);
}
int belong(int u) {
return fa[u] == u ? u : fa[u] = belong(fa[u]);
}

void path(int u) {
while(u) {
int v = match[pre[u]];
match[u] = pre[u];
match[pre[u]] = u;
u = v;
}
}
int lca(int u, int v) {
++tag;
u = belong(u);
v = belong(v);
while(vis[u] != tag) {
vis[u] = tag;
u = belong(pre[match[u]]);
if(v) swap(u, v);
}
return u;
}
void connect(int u, int v, int root) {
while(belong(u) != root) {
pre[u] = v;
v = match[u];
if(flag[v] == 2) {
q.push(v);
flag[v] = 1;
}
if(belong(u) == u) fa[u] = root;
if(belong(v) == v) fa[v] = root;
u = pre[v];
}
}
bool bfs(int u) {

memset(flag, 0, sizeof flag);
memset(pre, 0, sizeof pre);
for(int i=1; i<=n; ++i) fa[i] = i;

clear(q);
q.push(u);
flag[u] = 1;

while(!q.empty()) {
u = q.front();
q.pop();
for(int v=1; v<=n; ++v) {
if(!g[u][v]) continue;
if(flag[v] == 0) {
pre[v] = u;
if(match[v] == 0) {
path(v);
return true;
}
q.push(match[v]);
flag[v] = 2;
flag[match[v]] = 1;
} else {
if(flag[v] == 2) continue;
if(belong(u) == belong(v)) continue;
int root = lca(u, v);
connect(u, v, root);
connect(v, u, root);
}
}
}
return false;
}
int main() {

scanf("%d%d", &n, &m);
for(int i=1, u, v; i<=m; ++i) {
scanf("%d%d", &u, &v);
g[u][v] = g[v][u] = true;
}

for(int i=1; i<=n; ++i)
if(match[i] == 0 and bfs(i)) ++ans;

printf("%d\n", ans);
for(int i=1; i<=n; ++i) printf("%d ", match[i]);

return 0;
}

UOJ80 二分图最大权匹配(BFS - KM)

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
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

typedef long long LL;
const int N = 440;
const LL INF = 0x3f3f3f3f3f3f3f3f;

bool vis[N];
int pre[N], link[N], res[N];
int n, m, e;
LL g[N][N], lx[N], ly[N], d[N], ans = 0;

void bfs(int k) {

int x, y = 0;
LL min1 = INF, delta;
memset(vis, false, sizeof vis);
memset(pre, 0, sizeof pre);
memset(d, 0x3f, sizeof d);
link[y] = k;

do {
x = link[y], delta = INF, vis[y] = true;
for(int i=1; i<=m; ++i) {
if(!vis[i]) {
if(d[i] > lx[x] + ly[i] - g[x][i]) {
d[i] = lx[x] + ly[i] - g[x][i];
pre[i] = y;
}
if(delta > d[i]) delta = d[i], min1 = i;
}
}
for(int i=0; i<=m; ++i)
if(vis[i]) lx[link[i]] -= delta, ly[i] += delta;
else d[i] -= delta;
y = min1;
} while(link[y]);
while(y) link[y] = link[pre[y]], y = pre[y];
}

int main() {

scanf("%d%d%d", &n, &m, &e);
m = max(n, m);
for(int i=1; i<=e; ++i) {
int u, v; LL w;
scanf("%d%d%lld", &u, &v, &w);
g[u][v] = w;
lx[u] = max(lx[u], w);
}

for(int i=1; i<=n; ++i) bfs(i);

for(int i=1; i<=m; ++i) {
if(!g[link[i]][i]) continue;
ans += g[link[i]][i];
res[link[i]] = i;
}
printf("%lld\n", ans);
for(int i=1; i<=n; ++i) printf("%d ", res[i]);

return 0;
}
发布于

2020-04-19

更新于

2023-01-11

许可协议

评论