匹配:模板

UOJ78 二分图最大匹配(DFS - KM)

#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)

#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)

#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;
}