BZOJ4435——[Cerc2015]Juice Junctions

0、题目大意：求两点之间的最小割之和

1、分析：很明显，最小割树，我们发现这个题并不能用n^3的方法来求答案。。

所以我们记录下所有的边，然后把边从大到小排序，然后跑一边类似kruskal的东西，顺便统计答案

TAT，这个题我被卡常数了，贴上TLE的代码吧。。。

#include <queue>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define inf 214748364

struct Edge{
    int from, to, cap, flow, next;
};
int head[3010], cur[3010];
Edge G[20010];
int tot;
int d[3010];
bool vis[3010];
int s, t, n, m;
int a[3010];
int b[3010];

inline void init(){
    memset(head, -1, sizeof(head));
    tot = -1;
    return;
}

inline void insert(int from, int to, int cap){
    G[++ tot] = (Edge){from, to, cap, 0, head[from]};
    head[from] = tot;
    G[++ tot] = (Edge){to, from, 0, 0, head[to]};
    head[to] = tot;
    return;
}

inline bool BFS(){
    for(int i = 1; i <= n; i ++) vis[i] = 0;
    queue<int> Q;
    Q.push(s);
    vis[s]=1;
    d[s]=0;
    while(!Q.empty()){
        int x = Q.front(); Q.pop();
        for(int i = head[x]; i != -1; i = G[i].next){
            Edge& e = G[i];
            if(e.cap - e.flow > 0 && !vis[e.to]){
                vis[e.to] = 1;
                d[e.to]=d[x]+1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

inline int dfs(int x, int a){
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i != -1; i = G[i].next){
        Edge& e = G[i];
        if(d[x]+1 == d[e.to] && (f = dfs(e.to, min(e.cap - e.flow, a))) > 0){
            e.flow += f;
            G[i ^ 1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}

inline int maxflow(){
    int res = 0;
    while(BFS()){
        for(int i = 1; i <= n; i ++) cur[i] = head[i];
        res += dfs(s, inf);
    }
    return res;
}

inline void Clear(){
    for(int i = 0; i <= tot; i += 2){
        G[i].flow = G[i ^ 1].flow = (G[i].flow + G[i ^ 1].flow) / 2;
    }
}

struct node{
    int u, v, w;
    inline bool operator < (const node& rhs) const{
        return w > rhs.w;
    }
} T[20010];
int num;

inline void add(int u, int v, int w){
    T[++ num].u = u;
    T[num].v = v;
    T[num].w = w;
}

inline void solve(int l, int r){
    if(l == r) return;
    int rl = rand() % (r - l + 1) + l;
    int rr = rand() % (r - l + 1) + l;
    if(rl == rr) rl ++;
    if(rl > r) rl -= 2;
    s = a[rl], t = a[rr];
    Clear();

    int tw = maxflow();
    //puts("fuck");
    add(a[rl], a[rr], tw);
    int L = l, R = r;
    for(int i = l; i <= r; i ++){
        if(vis[a[i]]) b[L ++] = a[i];
        else b[R --] = a[i];
    }
    for(int i = l; i <= r; i ++) a[i] = b[i];
    solve(l, L - 1); solve(L, r);
}

int ff[20010], size[20010];

inline int find(int x){
    return ff[x] == x ? x : ff[x] = find(ff[x]);
}

int main(){
   // srand(time(NULL));
    scanf("%d%d", &n, &m);
    init();
    for(int i = 1; i <= m; i ++){
        int u, v;
        scanf("%d%d", &u, &v);
        insert(u, v, 1); insert(v, u, 1);
    }
    for(int i = 1; i <= n; i ++) a[i] = i;
    solve(1, n);
    LL ret = 0;
    sort(T + 1, T + num + 1);
    for(int i = 1; i <= n; i ++) size[i] = 1, ff[i] = i;
   // puts("fuck");
    for(int i = 1; i <= num; i ++){
        int tx = find(T[i].u), ty = find(T[i].v);
        if(size[tx] < size[ty]) swap(tx, ty);
        ret += (LL)T[i].w * size[tx] * size[ty];
        ff[ty] = tx;
        size[tx] += size[ty];
    }
    printf("%lld\n", ret);
    return 0;
}