http://codeforces.com/gym/103104

I. Sequence

题意：mm 个限制条件：pindex≠valuep_{index}\neq value。问有多少个四元组 (A,B,L,R)(A,B,L,R) 满足 ∀i∈[A,B]\forall i\in [A,B]，pip_i 可以取 [L,R][L,R] 中任意数。1≤A,B,L,R≤N1 \le A,B,L,R \le N

单调栈

问题转化为二维网格上有 mm 个坏点，问有几个只包含好点的矩形。

dp[i][j]dp[i][j] 表示以 (i,j)(i,j) 为右下端点的矩形的个数。

遍历行，遍历列，对每一行维护一个递增的单调栈，左上端点的取法等于栈顶的左上位置取法数+当前位置到栈顶位置构成的矩形中的点数。

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#include <bits/stdc++.h>

bool dbg = true;
#define debug(x) if (dbg) cout << #x << ":\t" << (x) << endl;
using namespace std;
#define ll long long
#define ull unsigned ll
const int N = 5005;
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;
const ll inv2 = (mod + 1) / 2;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

int n, m;
int mz[N][N];
int a[N];
int st[N], tp;
int dp[N][N];

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)mz[i][j] = 1;
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        mz[x][y] = 0;
    }
    ll ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++)
            if (!mz[i][j])a[j] = 0;
            else a[j]++;
        tp = 0;
        for (int j = 1; j <= n; j++) {
            while (tp && a[st[tp]] > a[j])tp--;
            if (tp)dp[i][j] = dp[i][st[tp]];
            dp[i][j] += (j - st[tp]) * a[j];
            ans += dp[i][j];
            st[++tp] = j;
        }
    }
    printf("%lld\n", ans);
    return 0;
}