[2016-03-18][POJ][1733][Parity game]

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• 时间:2016-03-18 09:55:52 星期五

• 题目编号:[2016-03-18][POJ][1733][Parity game]

• 题目大意:给定若干的区间范围内的数字的和,问从哪句开始是错误的

• 分析:

  • 带权并查集
  • 区间长度高达1000000000显然不可能直接建立数组,但是发现询问只有5000次,也就是最多出现了5000*2个点,离散化就可以解决问题
    • relation[i] i为区间的左端点,fa[i]为区间的右端点,relation[i]维护(i,fa[i])这段区间的和的奇偶状态,0表示偶数,1表示奇数
    • f(a,b) f(b,c)表示区间的就状态,那么f(a,c) = (f(a,b) + f(b,c))%2;

#include <map>
#include <cstdio>
using namespace std;
typedef long long LL;
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))
const int maxn = 5000*2 + 100;
int fa[maxn],relation[maxn];
map<LL ,int > m;
void ini(){
        m.clear();
        FOR(i,0,maxn)    fa[i] = i;
}
int fnd(int x){
        if(x == fa[x])  return x;
        int tmp = fa[x];
        fa[x] = fnd(fa[x]);
        relation[x] = (relation[x] + relation[tmp])&1;
        return fa[x];
}
int uni(int x,int y,int type){
        int fax = fnd(x),fay = fnd(y);
        if(fax == fay)      return 0;      
        fa[fax] = fay;
        relation[fax] = (relation[x] + relation[y] + type)&1;
        return 1;
}
int main(){
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
        int len,n,cur = 0,ans = 0,flg = 1;
        LL u,v;
        char str[10];
        scanf("%d%d",&len,&n);
        ini();     
        FOR(i,0,n){
                if(flg){
                        scanf("%I64d%I64d%s",&u,&v,str);
                        --u;
                        if(!m.count(u)) u = m[u] = cur++;
                        else u = m[u];
                        if(!m.count(v)) v = m[v] = cur++;
                        else v = m[v];
                        int type = (str[0] == ‘o‘?1:0);
                        if(!uni(u,v,type)){
                                int t = (relation[ u ] + relation[ v ])&1;
                                if(t != type){
                                        flg = 0;
                                        continue;
                                }
                        }
                        ++ans;
                }else scanf("%*I64d%*I64d%*s");
        }
        printf("%d\n",ans);
        return 0;
}

[2016-03-18][POJ][1733][Parity game]

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原文地址：http://www.cnblogs.com/qhy285571052/p/1beb05cb78ccdde9544c55cf70360550.html