hdu 5517 Triple

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5517

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前置知识以及思维难度上不算很高 不过实现对于我这种弱渣来说还是太复杂了（很多处理比较绕）

我们稍作分析（xiangyun）后可以发现 有可能最优且本质不同的三元组C只有不超过M个

对于二元组A b相同时只用保留a最大的 并且把相同的A合并掉

处理完之后再将B与之合体 得到不超过M个C

此时先按一个元素排序 再根据另两个元素值 用二维树状数组统计下就好了

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#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, lim = 1001;
int t, n, m, len, lenm;
struct sa
{
    int x,y,c;
}a[N],ua[N];
struct sc
{
    int x,y,z,c;
}c[N];
bool f[lim][lim];
bool cmp1(const sa &aa,const sa &bb)
{
    return aa.y < bb.y || (aa.y == bb.y && aa.x > bb.x);
}
bool cmp2(const sc &aa,const sc &bb)
{
    return aa.x > bb.x || (aa.x == bb.x && aa.y < bb.y)
    || (aa.x == bb.x && aa.y == bb.y && aa.z < bb.z);
}
int main()
{
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i)
            scanf("%d%d", &a[i].x, &a[i].y);
        sort(a + 1, a + 1 + n, cmp1);
        len = lenm = 0;
        memset(ua, 0, sizeof ua);
        for(int i = 1; i <=n;)
        {
            ++len;
            ua[len].x = a[i].x;
            ua[len].y = a[i].y;
            int j = i;
            while(a[j + 1].x == a[i].x)
                ++j;
            ua[len].c = j - i + 1;
            while(a[j]. y == a[i].y)
                ++j;
            i = j;
        }
        int tx,ty,tz;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d%d%d", &tx, &ty, &tz);
            int L = 1, R = len, mid;
            while(L < R)
            {
                mid = (L + R) >> 1;
                if(ua[mid].y >= tz)
                    R = mid;
                else
                    L = mid + 1;
            }
            if(ua[R].y == tz)
            {
                ++lenm;
                c[lenm].x = ua[R].x;
                c[lenm].y = lim - tx;
                c[lenm].z = lim - ty;
                c[lenm].c = ua[R].c;
            }
        }
        n = lenm;
        sort(c + 1,c + 1 + n,cmp2);
        memset(f, 0, sizeof f);
        long long ans = 0;
        for(int i = 1; i <= n; ++i)
        {
            if(c[i].x == c[i + 1].x && c[i].y == c[i + 1].y &&
               c[i].z == c[i + 1].z)
            {
                c[i + 1].c += c[i].c;
                continue;
            }
            bool tmp = 0;
            for(int p = c[i].y; p ; p -=  p & -p)
                for(int q = c[i].z; q; q -= q & -q)
                    tmp |= f[p][q];
            if(!tmp)
                ans += c[i].c;
            for(int p = c[i].y; p < lim; p += p & -p)
                for(int q = c[i].z; q < lim; q += q & -q)
                    f[p][q] = 1;
        }
        printf("Case #%d: %lld
", ca, ans);
    }
    return 0;
}