HDOJ【5575】【单调栈】【可并堆】

来源：互联网发布：淘宝家具店铺排行榜编辑：程序博客网时间：2024/06/16 00:02
好久木有做题啦！
/* I will wait for you */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <algorithm>#include <iostream>#include <fstream>#include <vector>#include <queue>#include <deque>#include <set>#include <map>#include <string>#include <ext/pb_ds/priority_queue.hpp>#define fi first#define se secondusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> pii;typedef map<int, int> mii;const int maxn = 500010;const int maxm = 1010;const int maxs = 26;const int inf = 0x3f3f3f3f;const int P = 1000000007;const double eps = 1e-6;inline ll read(){    ll x = 0, f = 1; char ch = getchar();    while (ch < '0' || ch > '9')          f = (ch == '-' ? -1 : 1), ch = getchar();    while (ch >= '0' && ch <= '9')           x = x * 10 + ch - '0', ch = getchar();    return x * f;}struct query{    int x, y, type;} e[maxn];bool operator < (query a, query b){    return a.y > b.y;}int n, m, top, h[maxn], sta[maxn],     sum_1[maxn], sum_2[maxn], dp[maxn];__gnu_pbds::priority_queue<query> pri[maxn];void init(){    for (int i = 1; i <= n; i++) {        pri[i].clear(), dp[i] = 0;        sum_1[i] = sum_2[i] = 0;    }}int main(){    int T = read(), cnt = 0;        while (T--) {        n = read(), m = read(), init();                for (int i = 1; i <= n - 1; i++)            h[i] = read();        h[0] = h[n] = inf;                for (int i = 1; i <= m; i++) {            int x = read(), y = read() + 1, t = read();            e[i] = (query) {x, y, t};            pri[x].push(e[i]), sum_2[x] += (t == 0);        }                top = sta[0] = 0;                for (int i = 1; i <= n; i++) {                        while (h[i] > h[sta[top]]) {                int cnt_1 = 0, cnt_2 = 0, hig = 0;                int ans = sum_1[i] + sum_2[i], j = sta[top];                                while (!pri[i].empty() && pri[i].top().y <= h[j]) {                    query tmp = pri[i].top(); pri[i].pop();                    if (tmp.y != hig) {                        sum_1[i] += cnt_1, cnt_1 = 0;                        sum_2[i] -= cnt_2, cnt_2 = 0;                        ans = max(ans, sum_1[i] + sum_2[i]);                        hig = tmp.y;                    }                    cnt_1 += (tmp.type == 1);                    cnt_2 += (tmp.type == 0);                }                                sum_1[i] += cnt_1, sum_2[i] -= cnt_2;                ans = max(ans, sum_1[i] + sum_2[i]);                dp[i] = max(dp[i], dp[j] + ans);                                sum_1[i] += sum_1[j], sum_2[i] += sum_2[j];                top -= 1, pri[i].join(pri[j]);            }                        int cnt_1 = 0, cnt_2 = 0, hig = 0;            int ans = sum_1[i] + sum_2[i], j = sta[top];                        while (!pri[i].empty() && pri[i].top().y <= h[i]) {                query tmp = pri[i].top(); pri[i].pop();                if (tmp.y != hig) {                    sum_1[i] += cnt_1, cnt_1 = 0;                    sum_2[i] -= cnt_2, cnt_2 = 0;                    ans = max(ans, sum_1[i] + sum_2[i]);                    hig = tmp.y;                }                cnt_1 += (tmp.type == 1);                cnt_2 += (tmp.type == 0);            }                            sum_1[i] += cnt_1, sum_2[i] -= cnt_2;            ans = max(ans, sum_1[i] + sum_2[i]);            dp[i] = max(dp[i], dp[j] + ans);                            sta[++top] = i;        }                printf("Case #%d: %d\n", ++cnt, dp[n]);    }    return 0;}
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