HDU 3225 Farming

HDU_3225

    这个题目和求被覆盖K次的矩形面积的并比较相似，只要把dp的过程稍加修改，用len[i][j]表示第i个节点控制的区域中第j种蔬菜覆盖的长度，然后根据当前节点的状态和子节点的状态完成状态转移。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#define MAXD 60010

int N, K, M, ty[MAXD], cnt[4 * MAXD][3], len[4 * MAXD][3], price[5], hash[110];

struct Seg

{

    int x, y1, y2, col, p;

}seg[MAXD];

int cmpint(const void *_p, const void *_q)

{

    int *p = (int *)_p, *q = (int *)_q;

    return *p < *q ? -1 : 1;

}

int cmps(const void *_p, const void *_q)

{

    Seg *p = (Seg *)_p, *q = (Seg *)_q;

    return p->x < q->x ? -1 : 1;

}

void build(int cur, int x, int y)

{

    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;

    memset(cnt[cur], 0, sizeof(cnt[cur])), memset(len[cur], 0, sizeof(len[cur]));

    if(x == y)

        return ;

    build(ls, x, mid);

    build(rs, mid + 1, y);

}

void init()

{

    int i, j, k, p[5], x1, y1, x2, y2, s;

    scanf("%d%d", &N, &K);

    for(i = 0; i < K; i ++)

    {

        scanf("%d", &p[i]);

        price[i] = p[i];

    }

    qsort(price, K, sizeof(price[0]), cmpint);

    for(i = 0; i < K; i ++)

        hash[price[i]] = i;

    for(i = 0; i < N; i ++)

    {

        j = i << 1, k = (i << 1) | 1;

        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &s);

        -- s;

        seg[j].x = x1, seg[k].x = x2;

        seg[j].y1 = seg[k].y1 = y1, seg[j].y2 = seg[k].y2 = y2;

        seg[j].col = 1, seg[k].col = -1;

        seg[j].p = seg[k].p = hash[p[s]];

        ty[j] = y1, ty[k] = y2;

    }

    qsort(ty, N << 1, sizeof(ty[0]), cmpint);

    M = -1;

    for(i = 0; i < (N << 1); i ++)

        if(i == 0 || ty[i] != ty[i - 1])

            ty[++ M] = ty[i];

    build(1, 0, M - 1);

}

int BS(int x)

{

    int min = 0, max = M + 1, mid;

    for(;;)

    {

        mid = (min + max) >> 1;

        if(mid == min)

            break;

        if(ty[mid] <= x)

            min = mid;

        else

            max = mid;

    }

    return mid;

}

void update(int cur, int x, int y)

{

    int ls = cur << 1, rs = (cur << 1) | 1;

    memset(len[cur], 0, sizeof(len[cur]));

    if(x == y)

    {

        int i;

        for(i = K - 1; i >= 0; i --)

            if(cnt[cur][i])

            {

                len[cur][i] = ty[y + 1] - ty[x];

                break;

            }

    }

    else

    {

        int i, t = 0;

        for(i = K - 1; i >= 0; i --)

        {

            if(cnt[cur][i])

            {

                len[cur][i] = ty[y + 1] - ty[x] - t;

                break;

            }

            else

            {

                len[cur][i] = len[ls][i] + len[rs][i];

                t += len[cur][i];

            }

        }

    }

}

void refresh(int cur, int x, int y, int s, int t, int c, int p)

{

    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;

    if(x >= s && y <= t)

    {

        cnt[cur][p] += c;

        update(cur, x, y);

        return ;

    }

    if(mid >= s)

        refresh(ls, x, mid, s, t, c, p);

    if(mid + 1 <= t)

        refresh(rs, mid + 1, y, s, t, c, p);

    update(cur, x, y);

}

void solve()

{

    int i, j, k;

    long long int ans = 0, t;

    qsort(seg, N << 1, sizeof(seg[0]), cmps);

    seg[N << 1].x = seg[(N << 1) - 1].x;

    for(i = 0; i < (N << 1); i ++)

    {

        j = BS(seg[i].y1), k = BS(seg[i].y2);

        if(j < k)

            refresh(1, 0, M - 1, j, k - 1, seg[i].col, seg[i].p);

        t = 0;

        for(j = 0; j < K; j ++)

            t += price[j] * len[1][j];

        ans += t * (seg[i + 1].x - seg[i].x);

    }

    printf("%I64d\n", ans);

}

int main()

{

    int t, tt;

    scanf("%d", &t);

    for(tt = 0; tt < t; tt ++)

    {

        init();

        printf("Case %d: ", tt + 1);

        solve();

    }

    return 0;

}