hdu 3622 Bomb Game （二分+2-sat）

Bomb Game

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5716    Accepted Submission(s): 2069

Problem Description

Robbie is playing an interesting computer game. The game field is an unbounded 2-dimensional region. There are N rounds in the game. At each round, the computer will give Robbie two places, and Robbie should choose one of them to put a bomb. The explosion area of the bomb is a circle whose center is just the chosen place. Robbie can control the power of the bomb, that is, he can control the radius of each circle. A strange requirement is that there should be no common area for any two circles. The final score is the minimum radius of all the N circles.
Robbie has cracked the game, and he has known all the candidate places of each round before the game starts. Now he wants to know the maximum score he can get with the optimal strategy.

 

Input

The first line of each test case is an integer N (2 <= N <= 100), indicating the number of rounds. Then N lines follow. The i-th line contains four integers x_1i, y_1i, x_2i, y_2i, indicating that the coordinates of the two candidate places of the i-th round are (x_1i, y_1i) and (x_2i, y_2i). All the coordinates are in the range [-10000, 10000].

 

Output

Output one float number for each test case, indicating the best possible score. The result should be rounded to two decimal places.

 

Sample Input

21 1 1 -1-1 -1 -1 121 1 -1 -11 -1 -1 1

 

Sample Output

1.411.00

 

Source

2010 Asia Regional Tianjin Site —— Online Contest

 

Recommend

题意：每一对炸弹只能选一个（明显2-sat），每个炸弹半径自定，爆炸范围不可
相交，求那个最小半径的最大值（每种策略的最小半径不同）。

思路： 2-sat一眼题吧， 二分半径， 2-sat判定， 2对炸弹有4种情况， 每种情况如果冲突，那就 i - j`, j - i`，就好了

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <stack>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 2e3 + 5;int n, m, low[maxn], dfn[maxn], id[maxn], scc_cnt, dfs_cnt;int x[maxn], y[maxn];vector<int> v[maxn];stack<int> s;void init(){    memset(low, 0, sizeof(low));    memset(id, 0, sizeof(id));    memset(dfn, 0, sizeof(dfn));    scc_cnt = dfs_cnt = 0;    for(int i = 0; i < maxn; i++)        v[i].clear();    while(!s.empty())        s.pop();}void addedge(int x, int y){    v[x].push_back(y);}void tarjan(int x){    dfn[x] = low[x] = ++dfs_cnt;    s.push(x);    for(int i = 0; i < v[x].size(); i++)    {        int to = v[x][i];        if(!dfn[to])        {            tarjan(to);            low[x] = min(low[x], low[to]);        }        else if(!id[to])            low[x] = min(low[x], dfn[to]);    }    if(low[x] == dfn[x])    {        scc_cnt++;        while(1)        {            int u = s.top();            s.pop();            id[u] = scc_cnt;            if(x == u) break;        }    }}void scc(){    for(int i = 1; i <= 2*n ; i++)        if(!dfn[i])            tarjan(i);}double dist(int i, int j){    return (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]);}int check(double r){    init();    for(int i = 1; i <= 2*n; i+=2)    {        for(int j = i+2; j <= 2*n; j+=2)        {            if(dist(i, j) < r*r*4)                addedge(i, j+1), addedge(j, i+1);            if(dist(i, j+1) < r*r*4)                addedge(i, j), addedge(j+1, i+1);            if(dist(i+1, j) < r*r*4)                addedge(i+1, j+1), addedge(j, i);            if(dist(i+1, j+1) < r*r*4)                addedge(i+1, j), addedge(j+1, i);        }    }    scc();    for(int i = 1; i <= 2*n; i+=2)        if(id[i] == id[i+1])            return 0;    return 1;}int main(){    while(~scanf("%d", &n))    {        for(int i = 1; i <= 2*n; i+=2)            scanf("%d%d%d%d", &x[i], &y[i], &x[i+1], &y[i+1]);        double l = 0, r = 1e9, mid;        int t = 300;        while(t--)        {            mid = (l+r)/2;            if(check(mid))                l = mid;            else                r = mid;        }        printf("%.2f\n", l);    }    return 0;}

Bomb Game

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5716    Accepted Submission(s): 2069

Problem Description

Robbie is playing an interesting computer game. The game field is an unbounded 2-dimensional region. There are N rounds in the game. At each round, the computer will give Robbie two places, and Robbie should choose one of them to put a bomb. The explosion area of the bomb is a circle whose center is just the chosen place. Robbie can control the power of the bomb, that is, he can control the radius of each circle. A strange requirement is that there should be no common area for any two circles. The final score is the minimum radius of all the N circles.
Robbie has cracked the game, and he has known all the candidate places of each round before the game starts. Now he wants to know the maximum score he can get with the optimal strategy.

 

Input

The first line of each test case is an integer N (2 <= N <= 100), indicating the number of rounds. Then N lines follow. The i-th line contains four integers x_1i, y_1i, x_2i, y_2i, indicating that the coordinates of the two candidate places of the i-th round are (x_1i, y_1i) and (x_2i, y_2i). All the coordinates are in the range [-10000, 10000].

 

Output

Output one float number for each test case, indicating the best possible score. The result should be rounded to two decimal places.

 

Sample Input

21 1 1 -1-1 -1 -1 121 1 -1 -11 -1 -1 1

 

Sample Output

1.411.00

 

Source

2010 Asia Regional Tianjin Site —— Online Contest

 

Recommend

lcy

 

Statistic | Submit | Discuss | Note