ZOJ-The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple-D

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ACM模版

描述

题解

离散化找公共区间，然后对区间进行一定规则的累加就行了。
具体规则就看翻译了，能翻译对，就 AC，翻译不对，就可以洗洗碎了……还好猜样例猜到了规则！

哎，英语渣渣打比赛真是累~~~全靠猜！

对了，这个题的数据有些迷，题目说好了 1≤x,y≤100，我开了 MAXN=110 却 WA 了，开到200才 AC！！！痛心疾首啊，难道又是我翻译出了问题？

代码

#include <iostream>#include <algorithm>#include <cstring>using namespace std;const int MAXN = 200;struct Node{    int l;    int r;} AtoB[MAXN], BtoA[MAXN];int pos[MAXN * 4], pos_[MAXN * 4], vis[MAXN * 4], vis_[MAXN * 4];int bs(int l, int r, int k){    while (l < r)    {        int m = (l + r) / 2;        if (k <= pos_[m])        {            r = m;        }        else        {            l = m + 1;        }    }    return l;}int main(){//    freopen("/Users/zyj/Desktop/input.txt", "r", stdin);    int T;    cin >> T;    while (T--)    {        memset(vis, 0, sizeof(vis));        memset(vis_, 0, sizeof(vis_));        int n, m, x, y;        cin >> n >> m >> x >> y;        int cnt = 0;        for (int i = 0; i < x; i++)        {            cin >> AtoB[i].l >> AtoB[i].r;            pos[cnt++] = AtoB[i].l;            pos[cnt++] = AtoB[i].r;        }        for (int i = 0; i < y; i++)        {            cin >> BtoA[i].l >> BtoA[i].r;            pos[cnt++] = BtoA[i].l;            pos[cnt++] = BtoA[i].r;        }        sort(pos, pos + cnt);        int cnt_ = 0;        for (int i = 1; i < cnt; i++)        {            if (pos[i] != pos[i - 1])            {                pos[++cnt_] = pos[i];            }        }        cnt = 0;        for (int i = 0; i <= cnt_; i++)        {            pos_[cnt++] = pos[i] * 2;            pos_[cnt++] = pos[i] * 2 + 1;        }        for (int i = 0; i < x; i++)        {            int l = bs(0, cnt, AtoB[i].l * 2);            int r = bs(0, cnt, AtoB[i].r * 2);            for (int j = l; j <= r; j++)            {                vis[j] = 1;            }        }        for (int i = 0; i < y; i++)        {            int l = bs(0, cnt, BtoA[i].l * 2);            int r = bs(0, cnt, BtoA[i].r * 2);            for (int j = l; j <= r; j++)            {                vis_[j] = 1;            }        }        for (int i = 0; i < cnt; i++)        {            vis[i] = min(vis[i], vis_[i]);        }        int ans = 0;        int i = 0;        while (i < cnt)        {            int j = i;            while (vis[i])            {                i++;            }            if ((pos_[i - 1] - pos_[j]) / 2 + 1 >= m)            {                ans += (pos_[i - 1] - pos_[j]) / 2 + 1 - m + 1;            }            i++;        }        cout << ans << endl;    }    return 0;}

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