UVA 1393 - Highways

最新推荐文章于 2019-04-25 16:50:00 发布

Killerfear

最新推荐文章于 2019-04-25 16:50:00 发布

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分类专栏：数学：计数

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本文介绍了一个关于在Hackerland中规划高速公路的问题，该国由一个m×n的城市网格组成，城市间通过东西向和南北向的道路相连。任务是计算在现有道路的基础上，还需要建造多少条直线高速公路来直接连接任意两个城市。

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Hackerland is a happy democratic country with m×n cities, arranged in a rectangular m by n grid and connected by m roads in the east-west direction and n roads in the north-south direction. By public demand, this orthogonal road system is to be supplemented by a system of highways in sucha way that there will be a direct connection between any pair of cities. Each highway is a straight line going through two or more cities. If two cities lie on the same highway, then they are directly connected.If two cities are in the same row or column, then they are already connected by the existing orthogonal road system (each east-west road connects all the m cities in that row and each north-south road connects all the n cities in that column), thus no new highway is needed to connect them. Your task is to count the number of highway that has to be built (a highway that goes through several cities on a straight line is counted as a single highway).

$\epsfbox{p3720.eps}$

2 4
3 3
0 0

Sample Output

12
14

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <stdio.h>
#include <cmath>
#include <math.h>
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,b,a) for(int i = (b); i >= (a); --i)
#define clr(a,x) memset(a,(x),sizeof(a))
#define LL long long
#define eps 1e-10
using namespace std;
const int maxn = 300 + 5;

vector<int> v[maxn];
int num[maxn][maxn];
LL dp[maxn][maxn];

int gcd(int a,int b)
{
    while (a && b) {
        if (a > b) a %= b;
        else b %= a;
    }
    return a + b;
}

void pre_init()
{
    rep(i,1,maxn) {
        rep(j,1,maxn) if (gcd(i,j) == 1) {
            v[i].push_back(j);
            num[i][j] = 1;
        }
        rep(j,1,maxn) num[i][j] += num[i][j-1];
    }

    rep(j,1,maxn) rep(i,1,maxn) {
        num[i][j] += num[i-1][j];
    }
    clr(dp,0);
    rep(i,1,maxn)
    rep(j,1,maxn-1) {
        dp[i][j+1] = dp[i][j];
        rep(k,1,i+1) {
            dp[i][j+1] += num[k-1][j];
            dp[i][j+1] -= num[j/2][(k-1)/2];
            dp[i][j+1] += num[i-k][j];
            dp[i][j+1] -= num[(i-k)/2][j/2];
        }
    }
}

int main()
{
    #ifdef ACM
        freopen("in.txt", "r", stdin);
       // freopen("out.txt","w",stdout);
    #endif // ACM
    pre_init();
    int n,m;
    while (scanf("%d%d",&n,&m),n+m) {
        printf("%lld\n",dp[n][m]);
    }
}