2020.7.15
今天先開個(gè)小差，這圖書館接近零下的氣溫給爺凍傻了，這還咋寫題？馬上回去了，練一練網(wǎng)絡(luò)和tarjan吧。

這道題很簡(jiǎn)單，問有2n個(gè)人，n排座位，每個(gè)人都有喜歡的兩個(gè)座位，最多能安置多少個(gè)人？這不就是二分圖嘛，切了它！首先看到一排有兩個(gè)座位，換句話說這個(gè)點(diǎn)具有2對(duì)1的屬性，要素察覺，我們用拆點(diǎn)處理一下，每一排座位分為出邊和入邊兩個(gè)流量，入到出流量為2，這樣就可以一一對(duì)應(yīng)了，然后跑一邊最大流就ok了。好消息是AC了，悲劇是我發(fā)現(xiàn)我好像想多了，我把拆點(diǎn)這個(gè)步驟去掉好像也能AC，哭了，才想起來好像左右沒什么關(guān)系，如果分為左和右那我的方法可能管用？但好像并不需要的亞子。
令人吐血的AC記錄：

代碼（拆點(diǎn)）：

#include <bits/stdc++.h> using namespace std; #define limit (100000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步兩步 #define EPS 1e-6 #define FASTIO ios::sync_with_stdio(false);cin.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; inline ll read(){ll sign = 1, x = 0;char s = getchar();while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}return x * sign; }//快讀 inline void write(ll x){if(x < 0) putchar('-'),x = -x;if(x / 10) write(x / 10);putchar(x % 10 + '0'); } int n,m,vs,ve,p; int layer[limit],head[limit], cnt; struct node{int to ,next;ll flow, w; }edge[limit]; ll max_flow; void add_one(int u , int v, ll flow = 0){edge[cnt].to = v;edge[cnt].next = head[u];edge[cnt].flow = flow;edge[cnt].w = 0;head[u] = cnt++; } inline void add(int u, int v, ll flow){add_one(u,v,flow);add_one(v, u,0); } inline void init(bool flag = true){if(flag){memset(head, -1, sizeof(head));cnt = 0;}else{memset(layer, -1, sizeof(layer));} } inline bool bfs(){init(false);queue<int>q;layer[vs] = 0;//從第0層開始q.push(vs);while (q.size()){int u = q.front();q.pop();traverse(u){int v = edge[i].to,flow = edge[i].flow;if(layer[v] == -1 && flow > 0){layer[v] = layer[u] + 1;//迭代加深q.push(v);}}}return ~layer[ve]; } ll dfs(int u, ll flow){if(u == ve)return flow;ll rev_flow = 0,min_flow;traverse(u){int v =edge[i].to;ll t_flow = edge[i].flow;if(layer[v] == layer[u] + 1 && t_flow > 0){min_flow = dfs(v, min(flow, t_flow));flow -= min_flow;edge[i].flow -= min_flow;rev_flow += min_flow;edge[i^1].flow += min_flow;if(!flow)break;}}if(!rev_flow)layer[u] = -1;return rev_flow; } void dinic(){while (bfs()){max_flow += dfs(vs,inf);} } int main() { #ifdef LOCALFOPEN; #endifinit();n = read();vs = 50001, ve = vs + 1;//將座位拆成兩個(gè)點(diǎn)rep(i ,1,n){add(i + n * 2 , i + n * 3, 2);//每個(gè)點(diǎn)之間有兩個(gè)流量}rep(i ,1,n * 2){add(vs, i, 1);//本位置心相連int x = read(), y = read();add(i, n * 2 + x, 1);add(i, n * 2 + y, 1);}//拆點(diǎn)rep(i ,1, n){add(n * 3 + i, ve,2);}dinic();write(max_flow);return 0; }

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