2017 Mid Central Regional G. Faulty Robot(dfs + 尬模)
这道题看上去太像tarjan缩点了,我一上去本来想把所有的环给缩掉然后统计答案,后来发现哦,这道题不是这么回事儿。
给出黑边红边,一次性走至多只能走一次黑边,问有多少个点可以走到,并且让机器人停下来,就一次这也不是很难么,但是如果k >= 2就很难说了
爆搜就完事儿了,遇到黑边改一次再走,如果一个点走不通了就统计一下,记得入点要特判
#include <bits/stdc++.h> using namespace std; #define limit (20000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #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) #define debug(x) cout<<x<<endl 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; }//快读 void write(ll x){ if(x < 0) putchar('-'),x = -x; if(x / 10) write(x / 10); putchar(x % 10 + '0'); } int n; int head[limit],cnt; struct node{ int to,next; bool forced; }edge[limit]; void init(){ memset(head, -1, sizeof(head)); cnt = 0; } int ans; void add(int x, int y,bool flag = false){ edge[cnt].to = y; edge[cnt].next = head[x]; edge[cnt].forced = flag; head[x] = cnt++; } int vis[limit],tot[limit]; void dfs(int u, int flag){ bool ok = false; traverse(u){ int v = edge[i].to; bool status = edge[i].forced; if(status){ if(!vis[v]){ vis[v] = 1; dfs(v,flag); vis[v] = 0; } ok = true; }else if(flag){ if(!vis[v]){ vis[v] = 1; dfs(v,!flag); vis[v] = 1; } } } if(!ok)tot[u]= 1; } int main() { #ifdef LOCAL FOPEN; #endif n = read(); int m = read(); init(); rep(i ,1,m){ int x = read(), y =read(); if(x < 0){ add(-x,y, true); }else{ add(x,y); } } int ok = 1; traverse(1)if(edge[i].forced){ ok = 0; break; } tot[1] = ok; //vis[1] = 1;; ans = 0; dfs(1,true); rep(i,1,n){ ans += tot[i]; } write(ans); return 0; }
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