luogu P2765 魔术球问题

Posted 2021-03-11 gredcomet

题目中没有说球的上限是多少，只告诉了柱子，那么我们就应该以柱子为界去增加球，考虑将每两个能组成完全平方数的点连边，就形成了一个DAG(有向无环图)，由于是DAG，可以转换为最小覆盖问题，即最多有n条路径(柱子数)，求其能覆盖的最大点数，最小覆盖路径 = 节点数 - 最大匹配数，可以将其拆成二分图跑匈牙利/最大流，由Hall定理，|S| <= |T|，此处的|S|就等于节点数-最大匹配数，而|T|等于最小覆盖路径，就是柱子数n(n个路径必有n个节点)，在满足条件的情况下增加球的数量即可。

在求最小覆盖路径时，可以用匈牙利/dinic,首先都要先拆分成二分图，每个点拆成一个入点一个出点，用dinic时，就要再设一个源点一个汇点，进行多次增广路，每一次的flow就是其增加节点后增加的匹配数，而用匈牙利时可以不实际拆图，对每个节点都跑一次匈牙利即可。

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, nex;//, cap, flow, nex;
} edges[maxm];

int head[maxm], cur[maxm], cnt, level[maxm], pre[maxm];
bool vis[maxm];

void init() {
    memset(head, -1, sizeof(head));memset(pre, -1, sizeof(pre));
}

void addedge(int u, int v, int cap) {
    edges[cnt] = edge{u, v, head[u]};//cap, 0, head[u]};
    head[u] = cnt++;
}

bool dfs(int u) {
    for(int i = head[u]; i != -1; i = edges[i].nex) {
        int v = edges[i].v;
        if(!vis[v]) {
            vis[v] = true;
            if(pre[v] == -1|| dfs(pre[v])) {
                pre[v] = u;
                return true;
            }
        }
    }
    return false;
}

void run_case() {
    init();
    int n; cin >> n;
    int flow = 0, num = 0;
    while(num - flow <= n) {
        num++;
        for(int i = sqrt(num)+1; i*i<(num<<1); ++i) {
            //和dinic一样 由新加的点向原点连，保证能更新答案，因为从新点跑匈牙利
            addedge(num, i*i-num,0);
        }
        flow += dfs(num); memset(vis, 0, sizeof(vis));
    }
    cout << --num;
    for(int i = 1; i <= num; ++i) {
        if(!vis[i]) {
            cout << "
";
            for(int u = i; u!=-1; u = pre[u]) {
                vis[u] = true;
                cout << u << " ";
            }
                
        }
    }
}


int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    //cout.flush();
    return 0;
}

匈牙利

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, cap, flow, nex;
} edges[maxm];

int head[maxm], cur[maxm], cnt, level[maxm], pre[maxm];
bool vis[maxm];

void init() {
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap) {
    edges[cnt] = edge{u, v, cap, 0, head[u]};
    head[u] = cnt++;
}

void bfs(int s) {
    memset(level, -1, sizeof(level));
    queue<int> q;
    level[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edges[i].nex) {
            edge& now = edges[i];
            if(now.cap > now.flow && level[now.v] < 0) {
                level[now.v] = level[u] + 1;
                q.push(now.v);
            }
        }
    }
}

int dfs(int u, int t, int f) {
    if(u == t) return f;
    for(int& i = cur[u]; i != -1; i = edges[i].nex) {
        edge& now = edges[i];
        if(now.cap > now.flow && level[u] < level[now.v]) {
            int d = dfs(now.v, t, min(f, now.cap - now.flow));
            if(d > 0) {
                now.flow += d;
                edges[i^1].flow -= d;
                pre[u>>1] = now.v>>1;
                return d;
            }

        }
    }
    return 0;
}

int dinic(int s, int t) {
    int maxflow = 0;
    for(;;) {
        bfs(s);
        if(level[t] < 0) break;
        memcpy(cur, head, sizeof(head));
        int f;
        while((f = dfs(s, t, INF)) > 0)
            maxflow += f;
    }
    return maxflow;
}

void run_case() {
    init();
    int n; cin >> n;
    int s = 0, t = 1e5;
    int flow = 0, num = 0;
    while(num - flow <= n) {
        num++;
        addedge(s, num<<1, 1), addedge(num<<1, s, 0);
        addedge((num<<1)|1, t, 1), addedge(t, (num<<1)|1, 0);
        for(int i = sqrt(num)+1; i*i<(num<<1); ++i) {
            //是将原有的球的入点向新加的球的出点连边，保证能增加流
            addedge((i*i-num)<<1, (num<<1)|1, 1), addedge((num<<1)|1, (i*i-num)<<1, 0);
        }
        flow += dinic(s, t);
    }
    cout << --num;
    for(int i = 1; i <= num; ++i) {
        if(!vis[i]) {
            cout << "
";
            for(int u = i; u&&u!=t>>1; u = pre[u]) {
                vis[u] = true;
                cout << u << " ";
            }
                
        }
    }
}


int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    //cout.flush();
    return 0;
}

Dinic