POJ 1733 Parity game hash离散化+带权并查集

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由于数据范围过大，所以需要进行hash离散处理，然后就是建立并查集，ranks[x]表示节点x到根节点奇偶性，然后每次更新一次权值就好了，输出第一次错的位置，直接break就好了。

#include<iostream>#include<cstdio>#include<cstring>#include<ctime>#include<algorithm>#include<cstdlib>#include<cmath>#include<set>#include<bitset>#include<map>#include<stack>#include<queue>#include<vector>#include<utility>#define INF 0x3f3f3f3f#define inf 2*0x3f3f3f3f#define llinf 1000000000000000000#define pi acos(-1.0)#define mod 1000000007#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt rt<<1#define rrt rt<<1|1#define rep(i,a,b) for(int i=(a);i<(b);i++)#define per(i,a,b) for(int i=(b)-1;i>=(a);i--)#define mem(a,b) memset(a,b,sizeof(a))#define lb(x) (x&-x)#define gi(x) scanf("%d",&x)#define gi2(x,y) scanf("%d%d",&x,&y)#define gll(x) scanf("%lld",&x)#define gll2(x,y) scanf("%lld%lld",&x,&y)#define gc(x) scanf("%c",&x)using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int,int>P;/***********************************************/int par[10005],ranks[10005],Hash[10005],x,y,ss,l,n,up=1;string s;struct node{    int a,b,num;}qu[5005];int finds(int x){    if(par[x]==-1)return x;    int t=finds(par[x]);    ranks[x]=(ranks[x]+ranks[par[x]])%2;    return par[x]=t;}int H(int key,int n){    int l=1,r=n-1;    while(l<=r)    {        int m=(r+l)/2;        if(Hash[m]==key)return m;        else if(Hash[m]>key)r=m-1;        else l=m+1;    }}int main(){    gi2(l,n);    mem(par,-1);    rep(i,0,n)    {        gi2(x,y);cin>>s;x--;        qu[i].a=x;qu[i].b=y;        if(s[0]=='e')qu[i].num=0;        else qu[i].num=1;        Hash[up++]=x;Hash[up++]=y;    }    sort(Hash,Hash+up);    int uu=2;    rep(i,2,up)    {        if(Hash[i]!=Hash[i-1])Hash[uu++]=Hash[i];    }    int i=0;    for(;i<n;i++)    {        int x=H(qu[i].a,uu),y=H(qu[i].b,uu);        int xx=finds(x),yy=finds(y);        if(xx!=yy)        {            par[yy]=xx;            ranks[yy]=(ranks[x]+ranks[y]+qu[i].num)%2;        }        else if(((ranks[x]+ranks[y])&1)!=qu[i].num)break;    }    cout<<i<<endl;    return 0;}

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