网络流24题

网络流24题

洛谷题单：网络流24题

 

第一题：飞行员配对问题

二分图模板题，匈牙利可以过，并且匈牙利算法可以记录匹配的信息

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 100010
using namespace std;
int cnt, n, m, head[N], nxt[N], to[N], p[N];
bool vis[N];
inline int read() {
    int x = 0, y = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')
            y = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = x * 10 + c - '0', c = getchar();
    return x * y;
}
void add(int x, int y) {
    nxt[++cnt] = head[x];
    head[x] = cnt;
    to[cnt] = y;
}
bool match(int u) {
    for (int i = head[u]; ~i; i = nxt[i]) {
        int v = to[i];
        if (vis[v])
            continue;
        else
            vis[v] = true;
        if (p[v] == 0 || match(p[v])) {
            p[v] = u;
            return true;
        }
    }
    return false;
}
int pipei() {
    int ans = 0;
    for (int i = 1; i <= m; i++) {
        memset(vis, false, sizeof(vis));
        if (match(i))
            ans++;
    }
    return ans;
}
int main() {
    cnt = -1;
    memset(head, -1, sizeof(head)); memset(p,0,sizeof(p));
    m = read();
    n = read();
    int u, v;
    while (~scanf("%d%d", &u, &v) && u+v != -2) add(u, v);
    int k = pipei();
    printf("%d\n",k);
    for(int i = m+1; i <= n; i++)
        if(p[i] != 0)
            printf("%d %d\n",p[i],i);
    return 0;
}

 

 

 

第二题：太空飞行计划问题

最小割的应用，最大权闭合子图模板题，源点建正权，汇点建负权

最大权闭合子图的权值和 = max{被选择的点权和} = 正点权和−min{没被选择的正权点之和 + 被选择的负权点绝对值和} 

= 正点权和−最小割

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#define N 10010
using namespace std;
vector<int> g[N];
int cnt,n,m,s,t,head[N],nxt[N],to[N],d[N],cur[N],w[N],f[N];
const int INF = 0x3f3f3f3f;
void add(int x, int y, int z)
{
    nxt[++cnt] = head[x];
    head[x] = cnt;
    to[cnt] = y;
    w[cnt] = z;
}
bool bfs(int s, int t)
{
    memset(d,-1,sizeof(d));
    d[s] = 0;
    queue<int> q;
    q.push(s);
    while(!q.empty())
    {
        int p = q.front(); q.pop();
        for(int i = head[p]; ~i; i = nxt[i])
        {
            int v = to[i];
            if(d[v] == -1 && w[i])
            {
                d[v] = d[p]+1;
                q.push(v);
            }
        }
    }
    return d[t] != -1;
}
int dfs(int s, int flow)
{
    if(s == t) return flow;
    int ans = 0;
    for(int i = cur[s]; ~i; i = nxt[i])
    {
        int v = to[i];
        cur[s] = i;
        if(d[v] == d[s]+1 && w[i])
        {
            int dis = dfs(v,min(flow,w[i]));
            if(dis)
            {
                w[i] -= dis;
                w[i^1] += dis;
                flow -= dis;
                ans += dis;
                if(flow == 0) break;
            }
        }
    }
    return ans;
}
int dinic(int s, int t)
{
    int ans = 0;
    while(bfs(s,t))
    {
        memcpy(cur,head,sizeof(head));
        ans += dfs(s,INF);
    }
    return ans;
}
int main()
{
    int sum = 0;
    cnt = -1; memset(head,-1,sizeof(head));
    scanf("%d%d",&m,&n);
    s = 0; t = n+m+1;
    for(int i = 1; i <= m; i++)
    {
        int x; char c;
        scanf("%d",&x);
        sum += x;
        add(s,i+n,x); add(i+n,s,0);
        while(true)
        {
            scanf("%d%c",&x,&c);
            add(i+n,x,INF); add(x,i+n,0);
            if(c == '\n' || c == '\r') break;
        }
    }
    for(int i = 1; i <= n; i++)
    {
        int x;
        scanf("%d",&x);
        add(i,t,x); add(t,i,0);
    }
    sum -= dinic(s,t);
    for(int i = 1; i <= m; i++)
        if(d[i+n] != -1)
            printf("%d ",i);
    puts("");
    for(int i = 1; i <= n; i++)
        if(d[i] != -1)
            printf("%d ",i);
    puts("");
    printf("%d\n",sum);
    return 0;
}

 

 

 

第三题：最小路径覆盖问题

最小路径覆盖，可看代码注释，拆点转化为二分图最大匹配

 

// 选出的路径最少 <=> 终点最少 <=> 二分图左侧的未匹配点最少 <=> 二分图匹配数最大
// 最少路径= 最少终点 = 总点数-最大匹配数 = n-maxflow
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 18010
using namespace std;
int cnt,n,m,head[N],nxt[N],to[N],p[N],p1[N];
bool vis[N];
inline int read()
{
    int x = 0, y = 1; char c = getchar(); while(c < '0' || c > '9') {if(c == '-') y = -1; c = getchar();} while(c >= '0' && c <= '9') x = x*10+c-'0', c = getchar(); return x*y;
}
void add(int x, int y)
{
    nxt[++cnt] = head[x];
    head[x] = cnt;
    to[cnt] = y;
}
bool match(int u)
{
    for(int i = head[u]; ~i; i = nxt[i])
    {
        int v = to[i];
        if(vis[v]) continue;
        else vis[v] = true;
        if(p[v] == 0 || match(p[v]))
        {
            p[v] = u;
            p1[u] = v;
            return true;
        }
    }
    return false;
}
int pipei()
{
    int ans = 0;
    for(int i = 1; i <= n; i++)
    {
        memset(vis,false,sizeof(vis));
        if(match(i))
            ans++;
    }
    return ans;
}
int main()
{
    cnt = -1; memset(head,-1,sizeof(head));
    memset(p,0,sizeof(p)); memset(p1,0,sizeof(p1));
    n = read(); m = read();
    for(int i = 1; i <= m; i++)
    {
        int u,v;
        u = read(); v = read();
        add(u,v+n);
    }
    int ans = n-pipei();
    for(int i = 0; i < ans; i++)
    {
        int j,pre;
        vector<int> g;
        for(j = 1; j <= n; j++)
        {
            if(p1[j] != 0)
            {
                g.push_back(j);
                break;
            }
        }
        while(p1[j] != 0) {pre = j; j = p1[j]-n; p1[pre] = 0; g.push_back(j);}
        for(int k = 0; k < g.size(); k++)
            printf("%d ",g[k]);
        puts("");
    }
    printf("%d",ans);
    return 0;
}