比赛-Round (11 Jul)

1. 小区划分

动规。题目略坑，通过样例发现，每个住户必须属于一个小区。

#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

double f[810][85], A[810], B[810];

int main()
{
    int i, j, k, N, K;
    scanf("%d%d", &N, &K);
    for (i = 1; i <= N; ++i)
        scanf("%lf", &A[i]), A[i] += A[i-1];
    for (i = 1; i <= N; ++i)
        scanf("%lf", &B[i]), B[i] += B[i-1];
    for (i = 1; i <= N; ++i)
        for (k = 1; k <= i && k <= K; ++k) {
            if (k == 1) { f[i][1] = fabs(A[i]-B[i]); continue; }
            for (j = k-1; j < i; ++j) {
                f[i][k] = max(f[i][k], f[j][k-1]+fabs((A[i]-A[j])-(B[i]-B[j])));
            }
        }
    printf("%.6lf
", f[N][K]);
    return 0;
}

2. 直线的交点

把各直线与两条平行线的交点处理一下就转换成了逆序对问题。看代码比较容易理解。

#include <cstdio>
#include <algorithm>

using namespace std;

const int _N = 120000;

typedef long long LL;

double K[_N], B[_N], X[_N], Y[_N], SX[_N], SY[_N], C[_N], GG[_N], k, a, b;
int N;

void Add(int k, int d)
{
    for (int i = k; i <= N; i += i & -i)
        C[i] += d;
    return;
}

LL GetSum(int k)
{
    LL tmp = 0;
    for (int i = k; i; i ^= i & -i)
        tmp += C[i];
    return tmp;
}

int main()
{
    int i;
    scanf("%lf%lf%lf%d", &k, &a, &b, &N);
    
    for (i = 1; i <= N; ++i) {
        scanf("%lf%lf", &K[i], &B[i]);
        X[i] = (B[i]-a)/(k-K[i]);
        SX[i] = X[i];
        Y[i] = (B[i]-b)/(k-K[i]);
        SY[i] = Y[i]; 
    }
    
    sort(SX+1, SX+1+N);
    sort(SY+1, SY+1+N);
    
    for (i = 1; i <= N; ++i) {
        LL tmpx = (LL)(lower_bound(SX+1, SX+1+N, X[i])-SX);
        LL tmpy = (LL)(lower_bound(SY+1, SY+1+N, Y[i])-SY);
        GG[tmpx] = tmpy;
    }
    
    
    LL ggans = 0;
    for (i = 1; i <= N; ++i) {
        ggans += GetSum(N)-GetSum(GG[i]);
        Add(GG[i], 1);
    }
    printf("%lld
", ggans);
    
    return 0;
}

3. 数三角形

各种计数推，只要能搞出来就好了。

#include <cstdio> 

typedef long long LL;

LL cnt[120000][2];

LL C_k_2(LL v) { return v >= 2 ? v*(v-1)/2 : 0; }

int main()
{
    LL N, M, i;
    scanf("%lld%lld", &N, &M);
    for (i = 1; i <= M; ++i) {
        LL x, y, w;
        scanf("%lld%lld%lld", &x, &y, &w);
        ++cnt[x][w-1], ++cnt[y][w-1];
    }
    LL X = 0, Y = 0;
    for (i = 1; i <= N; ++i) {
        LL cnt1 = cnt[i][0], cnt2 = cnt[i][1];
        LL cnt3 = N-1-cnt[i][0]-cnt[i][1];
        X += cnt1*cnt2+cnt2*cnt3+cnt1*cnt3;
        Y += C_k_2(cnt1)+C_k_2(cnt2)+C_k_2(cnt3);
    }
    printf("%lld
", X-2*Y);
    
    return 0;
}