bzoj 4823: [Cqoi2017]老C的方块 [最小割]

4823: [Cqoi2017]老C的方块

题意：

鬼畜方块游戏不解释...

有些特殊边，有些四个方块组成的图形，方块有代价，删掉一些方块使得没有图形，最小化代价。

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比较明显的最小割，一个图形中必须删掉一个方块。

我的想法是方块拆点然后用INF连起来。

但是你不能随便连啊，否则可能会出现一些原来没有的限制。

要找到一个连边的顺序！也就是如何分层

画一画发现是可以做到的，然后建图就行了。

我是一层一层建的...

然后一开始忘记考虑一种图形WA了两次...

总共花了3个多小时...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
typedef long long ll;
const int N = 2e5+5, M = 2e6+5, INF = 1e9+5;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int m, n, k, s, t;
struct meow{int x, y, w;} a[N];
map<int, int> g[100005];

struct edge{int v, ne, c, f;} e[M];
int cnt=1, h[N];
inline void ins(int u, int v, int c) { 
    e[++cnt] = (edge){v, h[u], c, 0}; h[u] = cnt;
    e[++cnt] = (edge){u, h[v], 0, 0}; h[v] = cnt;
}

int start(int x, int y) {
    if((x-2)%4 == 0) {
        if(!(y&1)) return 1;
        else return 0;
    }
    if((x+1)%4 == 0) {
        if(y&1) return 2;
        else return 0;
    }
    return 0;
}

inline void link(int u, int v) { ins(u+k, v, INF); }
int t1[N], t2[N];
void build() {
    s = 0; t = k+k+1;
    for(int i=1; i<=k; i++) ins(i, i+k, a[i].w);
    int v;
    int *mark = t1, *mark2 = t2;
    for(int i=1; i<=k; i++) {
        int x = a[i].x, y = a[i].y, p = start(x, y); 
        if(!p) continue;
        ins(s, i, INF);  //printf("start (%d, %d)  %d\n", x, y, p);
        if(p == 1) {
            if(g[x].count(y+1)) v = g[x][y+1], link(i, v), mark[v] = 1;
            if(g[x].count(y-1)) v = g[x][y-1], link(i, v), mark[v] = 1;
            if(g[x+1].count(y)) v = g[x+1][y], link(i, v), mark[v] = 2;
        } else {
            if(g[x].count(y+1)) v = g[x][y+1], link(i, v), mark[v] = 2;
            if(g[x].count(y-1)) v = g[x][y-1], link(i, v), mark[v] = 2;
            if(g[x-1].count(y)) v = g[x-1][y], link(i, v), mark[v] = 1;
        }
    }
    swap(mark, mark2);
    for(int i=1; i<=k; i++) if(mark2[i]) {
        int x = a[i].x, y = a[i].y, p = mark2[i]; mark2[i] = 0;
        if(p == 1 && g[x-1].count(y)) v = g[x-1][y], link(i, v), mark[v] = 1;
        if(p == 2 && g[x+1].count(y)) v = g[x+1][y], link(i, v), mark[v] = 2;
    }
    swap(mark, mark2);
    for(int i=1; i<=k; i++) if(mark2[i]) {
        int x = a[i].x, y = a[i].y, p = mark2[i]; mark2[i] = 0;
        if(g[x].count(y+1)) v = g[x][y+1], link(i, v), mark[v] = 1;
        if(g[x].count(y-1)) v = g[x][y-1], link(i, v), mark[v] = 1;
        if(p == 1 && g[x-1].count(y)) v = g[x-1][y], link(i, v), mark[v] = 1;
        if(p == 2 && g[x+1].count(y)) v = g[x+1][y], link(i, v), mark[v] = 1;
    }
    for(int i=1; i<=k; i++) if(mark[i]) ins(i+k, t, INF);
}

namespace mf {
    int q[N], head, tail, vis[N], d[N];
    bool bfs() {
        memset(vis, 0, sizeof(vis));
        head = tail = 1;
        q[tail++] = s; d[s] = 0; vis[s] = 1;
        while(head != tail) {
            int u = q[head++]; 
            for(int i=h[u];i;i=e[i].ne) 
                if(!vis[e[i].v] && e[i].c > e[i].f) {
                    int v = e[i].v;
                    vis[v] = 1; d[v] = d[u]+1;
                    q[tail++] = v;
                    if(v == t) return true;
                }
        }
        return false;
    }
    int cur[N];
    int dfs(int u, int a) {
        if(u == t || a == 0) return a;
        int flow = 0, f;
        for(int &i=cur[u];i;i=e[i].ne) 
            if(d[e[i].v] == d[u]+1 && (f = dfs(e[i].v, min(a, e[i].c - e[i].f)) ) >0 ) {
                flow += f;
                e[i].f += f;
                e[i^1].f -= f;
                a -= f;
                if(a == 0) break;
            }
        if(a) d[u] = -1;
        return flow;
    }
    int dinic() {
        int flow = 0;
        while(bfs()) { 
            for(int i=s; i<=t; i++) cur[i] = h[i];
            flow += dfs(s, INF); 
        }
        return flow;
    }
}

int main() {
    freopen("in", "r", stdin);
    m=read(); n=read(); k=read();
    for(int i=1; i<=k; i++) a[i].x = read(), a[i].y = read(), a[i].w = read(), g[a[i].x][a[i].y] = i;
    build();
    int ans = mf::dinic();
    printf("%d\n", ans);
}