NOIp2010 关押罪犯

Posted 2020-08-04

二分+2-SAT

先预处理出所有的v，然后离散化一下，在那个的基础上二分，对于每次二分出的值约束边权超过所二分出的边权的两点。

  1 //OJ 1322
  2 //by Cydiater
  3 //2015.8.26
  4 #include <iostream>
  5 #include <cstdio>
  6 #include <cstring>
  7 #include <queue>
  8 #include <map>
  9 #include <ctime>
 10 #include <string>
 11 #include <algorithm>
 12 #include <iomanip>
 13 #include <cstdlib>
 14 #include <cmath>
 15 using namespace std;
 16 #define ll long long
 17 #define up(i,j,n)       for(int i=j;i<=n;i++)
 18 #define down(i,j,n)     for(int i=j;i>=n;i--)
 19 const int MAXN=1e6+5;
 20 const int oo=0x3f3f3f3f;
 21 inline int read(){
 22       char ch=getchar();int x=0,f=1;
 23       while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
 24       while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 25       return x*f;
 26 }
 27 map<int,int>m;
 28 int N,LINK[MAXN],M,tmp[MAXN],top=0,cnt=0,num[MAXN],leftt,rightt,len=0,group[MAXN],group_num,dfn[MAXN],low[MAXN],stack[MAXN],node[MAXN][2],part=0,dfs_clock=0;
 29 bool vis[MAXN];
 30 struct edge{
 31       int y,next,x;
 32 }e[MAXN];
 33 struct Prison{
 34       int x,y,v;
 35 }a[MAXN];
 36 namespace solution{
 37       inline void insert(int x,int y){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;e[len].x=x;}
 38       void init(){
 39             N=read();M=read();
 40             tmp[++top]=0;
 41             up(i,1,M){
 42                   a[i].x=read();a[i].y=read();a[i].v=read();
 43                   tmp[++top]=a[i].v;
 44             }
 45             sort(tmp+1,tmp+top+1);
 46             up(i,1,top)if(!m[tmp[i]]){
 47                   m[tmp[i]]=++cnt;
 48                   num[cnt]=tmp[i];
 49             }
 50             up(i,1,N)up(j,0,1)node[i][j]=++part;
 51       }
 52       void tarjan(int node){
 53             dfn[node]=low[node]=++dfs_clock;
 54             vis[node]=1;stack[++top]=node;
 55             for(int i=LINK[node];i;i=e[i].next)
 56                   if(!dfn[e[i].y]){
 57                         tarjan(e[i].y);
 58                         low[node]=min(low[node],low[e[i].y]);
 59                   }else if(vis[e[i].y]) low[node]=min(low[node],dfn[e[i].y]);
 60             if(low[node]==dfn[node]){
 61                   int tmp;group_num++;
 62                   do{
 63                         tmp=stack[top--];
 64                         vis[tmp]=0;
 65                         group[tmp]=group_num;
 66                   }while(tmp!=node);
 67             }
 68       }
 69       bool check(int xx){
 70             len=0;top=0;dfs_clock=0;group_num=0;
 71             memset(LINK,0,sizeof(LINK));
 72             memset(dfn,0,sizeof(dfn));
 73             memset(vis,0,sizeof(vis));
 74             up(i,1,M)if(a[i].v>xx){
 75                   int x=a[i].x,y=a[i].y;
 76                   //cout<<x<<‘ ‘<<y<<endl;
 77                   insert(node[x][0],node[y][1]);
 78                   insert(node[y][1],node[x][0]);
 79                   insert(node[x][1],node[y][0]);
 80                   insert(node[y][0],node[x][1]);
 81             }
 82             up(i,1,N<<1)if(!dfn[i])tarjan(i);
 83             up(i,1,N)if(group[node[i][1]]==group[node[i][0]])return 0;
 84             return 1;
 85       }
 86       void slove(){
 87             leftt=1;rightt=cnt;
 88             while(leftt+1<rightt){
 89                   int mid=(leftt+rightt)>>1;
 90                   if(check(num[mid]))     rightt=mid;
 91                   else                    leftt=mid;
 92             }
 93       }
 94       void output(){
 95             if(check(num[leftt])){
 96                   cout<<num[leftt]<<endl;
 97             }else
 98                   cout<<num[rightt]<<endl;
 99       }
100 }
101 int main(){
102       //freopen("input.in","r",stdin);
103       using namespace solution;
104       init();
105       slove();
106       output();
107       return 0;
108 }

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