ZOJ 3717 HDU 3622 二分+2-SAT

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题意：给你N组气球，每组有2个气球，每组要取一个气球，问最后使得N个气球都不相交，则气球半径R最大是多少。

思路：直接二分半径，然后2-sat判可行性。SCC之后如果有两个同组的点在同一个强联通分量里，那么则不可行。

这道题注意最后的二分结束之后还要取三位小数，看取了之后是否还是符合情况的，最后的那个操作是看别人的。。我WA到死了。。

我感觉这题这里太坑了。。。

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <vector>#include <stack>#include <map>#include <iomanip>#define PI acos(-1.0)#define Max 2505#define inf 1<<28#define LL(x) ( x << 1 )#define RR(x) ( x << 1 | 1 )#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define PII pair<int,int>using namespace std;int t ;#define N 2005double x[N << 1] , y[N << 1] ,z[N << 1] ;struct kdq{    int e , next ;}ed[N * 100] ;int head[N] , num ;int dfn[N << 1] , low[N << 1] ,st[N << 1] , top , dp ,inst[N << 1] , ca ,belong[N << 1] ;void add(int s ,int e){    ed[num].e = e ;    ed[num].next = head[s] ;    head[s] = num ++ ;}void init(){    mem(head , -1) ;    num = 0 ;    mem(dfn ,0) ;    mem(low, 0) ;    mem(st ,0) ;    mem(inst ,0) ;    mem(belong ,0) ;    top = dp = ca = 0 ;}inline double getdis(int i ,int j){    return sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) + (z[i] - z[j]) * (z[i] - z[j])) ;}void tarjan(int now){    low[now] = dfn[now] = ++ dp ;    st[top ++] = now ;    inst[now] = 1 ;    for (int i = head[now] ; ~i ; i = ed[i].next ){        int e = ed[i].e ;        if(!dfn[e]){            tarjan(e) ;            low[now] = min(low[now] , low[e])  ;        }        else if(inst[e]){            low[now] = min(low[now] , dfn[e]) ;        }    }    if(low[now] == dfn[now]){        ca ++ ;        int xx ;        do{            xx = st[-- top] ;            belong[xx] = ca ;            inst[xx] = 0 ;        }while(xx != now) ;    }}void build(double mid){    init() ;    for (int i = 0 ; i < t ; i ++ ){        for (int j = i + 1 ; j  < t ; j ++ ){            if(getdis(LL(i),LL(j)) < mid){                add(LL(i) , LL(j) ^ 1) ;                add(LL(j) , LL(i) ^ 1) ;            }            if(getdis(LL(i) ,RR(j)) < mid){                add(LL(i) , RR(j) ^ 1) ;                add(RR(j) , LL(i) ^ 1) ;            }            if(getdis(RR(i) , LL(j)) < mid){                add(RR(i) , LL(j) ^ 1) ;                add(LL(j) , RR(i) ^ 1) ;            }            if(getdis(RR(i) , RR(j)) < mid){                add(RR(i) , RR(j) ^ 1) ;                add(RR(j) , RR(i) ^ 1) ;            }        }    }}int fuckit(){    for (int i = 0 ; i  < t << 1 ; i ++ ){        top = dp = 0 ;        if(!dfn[i])tarjan(i) ;    }    for (int i = 0 ; i < t ; i ++ ){        if(belong[LL(i)] == belong[RR(i)])return 0 ;    }    return 1 ;}int main() {    while(cin >> t){        for (int i = 0 ; i < t ;i ++ ){            cin >> x[LL(i)] >> y[LL(i)] >> z[LL(i)] ;            cin >> x[RR(i)] >> y[RR(i)] >> z[RR(i)] ;        }//        cout << getdis(0 ,1) << endl;        double l = 0 , r = 20000 ;        double mid ;        while(r - l > 1e-5){            mid = (l + r) / 2 ;            build(mid) ;            if(fuckit()){                l = mid ;            }            else r = mid ;        }        double ans = mid / 2 ;//直接输出 mid / 2 就WA到死。        char aa[222] ;//太恶心。        sprintf(aa ,"%.3f" , ans) ;        sscanf(aa , "%lf" ,&ans) ;        build(ans * 2 ) ;        if(!fuckit())ans -= 0.001 ;        printf("%.3f\n",ans) ;    }    return 0 ;}

HDU 3622

两道题其实完全是一样的，不过一题是3D，一题是2D，解法完全相同，不过这题二分之后不需要判可行性了。

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <vector>#include <stack>#include <map>#include <iomanip>#define PI acos(-1.0)#define Max 2505#define inf 1<<28#define LL(x) ( x << 1 )#define RR(x) ( x << 1 | 1 )#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define PII pair<int,int>using namespace std;#define N 105double x[N << 1] , y[N << 1] ;struct kdq{    int e , next ;}ed[N * 1000] ;int dfn[N << 1] ,low[N << 1] , belong[N << 1] ,st[N << 1] ,inst[N << 1] ,head[N << 1] ;int dp , top , ca , num , n ;inline double getdis(int i ,int j){    return sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j])) ;}void add(int s ,int e){    ed[num].e = e ;    ed[num].next = head[s] ;    head[s] = num ++ ;}void init(){    mem(dfn ,0) ;    mem(low ,0) ;    mem(st ,0) ;    mem(head,-1) ;    mem(belong ,0) ;    mem(inst ,0) ;    dp = top = ca = num = 0 ;}void build(double mid){    init() ;    for (int i = 0 ; i < n ; i ++ ){        for (int j = i + 1 ; j < n ; j ++ ){            if(getdis(LL(i) , LL(j)) < mid){                add(LL(i) , LL(j) ^ 1) ;                add(LL(j) , LL(i) ^ 1) ;            }            if(getdis(LL(i) , RR(j)) < mid){                add(LL(i) , RR(j) ^ 1) ;                add(RR(j) , LL(i) ^ 1) ;            }            if(getdis(RR(i) , LL(j)) < mid){                add(RR(i) , LL(j) ^ 1) ;                add(LL(j) , RR(i) ^ 1) ;            }            if(getdis(RR(i) , RR(j)) < mid){                add(RR(i) , RR(j) ^ 1) ;                add(RR(j) , RR(i) ^ 1) ;            }        }    }}void tarjan(int now){    dfn[now] = low[now] = ++ dp ;    st[top ++] = now ;    inst[now] = 1 ;    for (int i = head[now] ; ~i ; i = ed[i].next ){        int e = ed[i].e ;        if(!dfn[e]){            tarjan(e) ;            low[now] = min(low[now] , low[e]) ;        }        else if(inst[e]){            low[now] = min(low[now] , dfn[e]) ;        }    }    if(low[now] == dfn[now]){        ca ++ ;        int xx ;        do{            xx = st[-- top] ;            belong[xx] = ca ;            inst[xx] = 0 ;        }while(xx != now) ;    }}int doit(){    for (int i = 0 ; i < n << 1 ; i ++ )if(!dfn[i])tarjan(i) ;    for (int i = 0 ; i < n ; i ++ )if(belong[LL(i)] == belong[RR(i)])return 0 ;    return 1 ;}int main() {    while(cin >> n ){        for (int i = 0 ; i < n ; i ++ ){            cin >> x[LL(i)] >> y[LL(i)] ;            cin >> x[RR(i)] >> y[RR(i)] ;        }        double l = 0 , r = 30000 ,mid ;        while(r - l > 1e-4){            mid = (l + r) / 2 ;            build(mid) ;            if(doit())l = mid ;            else r = mid ;        }        printf("%.2f\n",mid / 2) ;    }    return 0 ;}