Description
Input
Output
Sample Input
4 4 1 2 2 2 3 1 3 4 1 4 1 2
Sample Output
3
HINT
Source
题意:有n个点,每个点最多一个加油站,然后m个条件,输入a,b,c代表a,b两个点有c个加油站,问能满足所有条件的情况下,加油站最少是几个
思路:由于有5秒,可以试着搜索
#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <math.h> #include <algorithm> using namespace std; #define ls 2*i #define rs 2*i+1 #define up(i,x,y) for(i=x;i<=y;i++) #define down(i,x,y) for(i=x;i>=y;i--) #define mem(a,x) memset(a,x,sizeof(a)) #define w(a) while(a) #define LL long long const double pi = acos(-1.0); #define Len 200005 #define mod 19999997 const int INF = 0x3f3f3f3f; #define exp 1e-6 typedef pair<int,int> PII; vector<int> mat[Len]; int n,m,col[Len],flag = 1; void dfs1(int u,int father,int c) { if(!col[u]) col[u]=c; else if(col[u]==-c) { flag = 0; return; } int i,len = mat[u].size(); for(i = 0; i<len; i++) { int v = mat[u][i]; if(v == father) continue; if(col[v] == -c) continue; dfs1(v,u,-c); } } PII dfs2(int u,int father,int c) { PII ret = make_pair(0,0); if(!col[u]) col[u]=c; else if(col[u]==-c) { flag = 0; return ret; } if(c==1) ret.first++; else ret.second++; int len = mat[u].size(); for(int i=0; i<len; i++) { int v=mat[u][i]; if(v==father) continue; if(col[v]==-c) continue; PII tem=dfs2(v,u,-c); ret.first+=tem.first; ret.second+=tem.second; } return ret; } int main() { int i,j,k,a,b,c; mem(col,0); scanf("%d%d",&n,&m); w(m--) { scanf("%d%d%d",&a,&b,&c); c--; if(c == 0) { mat[a].push_back(b); mat[b].push_back(a); } else//确定能染色的,1为染,-1为不染 { if(col[a]==0) col[a] = c; else if(col[a]==-c) flag = 0;//染色情况相悖 if(col[b]==0) col[b] = c; else if(col[b]==-c) flag = 0; } } if(!flag) { printf("impossible\n"); return 0; } else { up(i,1,n)//更新 { if(col[i]) dfs1(i,i,col[i]); if(!flag) { printf("impossible\n"); return 0; } } int bb=0,ww=0; up(i,1,n)//计算所有确定的染色情况 { if(!col[i]) continue; if(col[i]==1) bb++; else ww++; } int tt = 0; up(i,1,n)//不确定的,进行搜索 { if(col[i])continue; PII tem = dfs2(i,i,1); tt+=min(tem.first,tem.second); } if(!flag) { printf("impossible\n"); return 0; } printf("%d\n",bb+tt); } return 0; }
时间: 2024-10-23 17:49:24