[swustoj1739] 魔术球问题 （最大流，最小路径覆盖）

题目链接：https://www.oj.swust.edu.cn/problem/show/1739

从1开始枚举球的个数，每次从残余网络更新总流量，最小路径覆盖刚好大于n时ret-1便是最多球。

之后根据容量为0的边找回匹配边即可。

用x << 1和x << 1 | 1拆点 比较方便。

#include <bits/stdc++.h>
using namespace std;

typedef struct Edge {
    int u, v, w, next;
}Edge;

const int inf = 0x7f7f7f7f;
const int maxn = 100200;

int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
Edge dedge[maxn<<1];
// bool vis[maxn]; // 记录经过的点
int S, T, N;

void init() {
    memset(dhead, -1, sizeof(dhead));
    for(int i = 0; i < maxn; i++) dedge[i].next = -1;
    S = 0; cnt = 0;
}

void adde(int u, int v, int w, int c1=0) {
    dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 
    dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
    dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 
    dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
}

bool bfs(int s, int t, int n) {
    // memset(vis, 0, sizeof(vis));
    queue<int> q;
    for(int i = 0; i < n; i++) dd[i] = inf;
    dd[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        for(int i = dhead[u]; ~i; i = dedge[i].next) {
            if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
                dd[dedge[i].v] = dd[u] + 1;
                // vis[dedge[i].v] = 1;
                if(dedge[i].v == t) return 1;
                q.push(dedge[i].v);
            }
        }
    }
    return 0;
}

int dinic(int s, int t, int n) {
    int st[maxn], top;
    int u;
    int flow = 0;
    while(bfs(s, t, n)) {
        for(int i = 0; i < n; i++) cur[i] = dhead[i];
        u = s; top = 0;
        while(cur[s] != -1) {
            if(u == t) {
                int tp = inf;
                for(int i = top - 1; i >= 0; i--) {
                    tp = min(tp, dedge[st[i]].w);
                }
                flow += tp;
                for(int i = top - 1; i >= 0; i--) {
                    dedge[st[i]].w -= tp;
                    dedge[st[i] ^ 1].w += tp;
                    if(dedge[st[i]].w == 0) top = i;
                }
                u = dedge[st[top]].u;
            }
            else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
                st[top++] = cur[u];
                u = dedge[cur[u]].v;
            }
            else {
                while(u != s && cur[u] == -1) {
                    u = dedge[st[--top]].u;
                }
                cur[u] = dedge[cur[u]].next;
            }
        }
    }
    return flow;
}

int k, n;
int path[maxn];
bool vis[maxn];

bool ok(int x) {
    int y = (int)sqrt(x);
    return x == y * y;
}

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d", &n)) {
        init();
        S = 0, T = 10001, N = T + 1;
        int ret = 0;
        int flow = 0;
        while(1) {
            ret++;
            for(int i = 1; i < ret; i++) {
                if(ok(i+ret)) adde(i<<1, (ret<<1)|1, 1);
            }
            adde(S, ret<<1, 1);
            adde((ret<<1)|1, T, 1);
            flow += dinic(S, T, N);
            if(ret - flow > n) break;
        }
        memset(vis, 0, sizeof(vis));
        memset(path, -1, sizeof(path));
        printf("%d\n", ret-1);
        int q = 0;
        for(int i = 1; i < ret; i++) {
            for(int j = dhead[i<<1]; ~j; j=dedge[j].next) {
                if(!dedge[j].w) {
                    path[i] = dedge[j].v >> 1;
                    break;
                }
            }
        }
        for(int i = 1; i < ret; i++) {
            if(!vis[i]) {
                vis[i] = 1;
                printf("%d", i);
                int j = path[i];
                while(j != 0 && j != T) {
                    printf(" %d", j);
                    vis[j] = 1;
                    j = path[j];
                }
                printf("\n");
            }
        }
    }
    return 0;
}