hdu 4698 Counting（计数）

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题目链接：hdu 4698 Counting

解题思路

考虑不可行的四元组，用所有方案数减掉。将x按照坐标排序，保证某些x满足时，对应的y不满足，累加求和。

代码

#include <cstdio>#include <cstring>#include <set>#include <algorithm>using namespace std;typedef set<int>::iterator iter;typedef long long ll;const int maxn = 1005;const int mod = 1e9 + 7;int C(int x) { return 1LL * x * (x+1) / 2 % mod; }struct Point {    int x, y;    bool operator < (const Point& u) const { return x < u.x; }}P[maxn];int N, M, K;set<int> pos;void add(int y, ll& s) {    iter it = pos.upper_bound(y);    iter rf = it;    it--;    if (*it == y) return;    s = (s - C(*rf - *it - 1) + mod) % mod;    s = (s + C(*rf - y - 1) + C(y - *it - 1)) % mod;    pos.insert(y);}ll solve(int s) {    pos.clear();    pos.insert(0);    pos.insert(M+1);    ll ret = 0, sum = C(M), l = P[s].x - P[s-1].x;    add(P[s].y, sum);    for (int i = s+1; i <= K; i++) {        if (P[i].x != P[i-1].x) {            int r = P[i].x - P[i-1].x;            ret = (ret + sum * l % mod * r % mod) % mod;        }        add(P[i].y, sum);    }    int r = N + 1 - P[K].x;    ret = (ret + sum * l % mod * r % mod) % mod;    return ret;}int main () {    while (scanf("%d%d%d", &N, &M, &K) == 3) {        for (int i = 1; i <= K; i++) scanf("%d%d", &P[i].x, &P[i].y);        P[0].x = 0, P[K+1].x = N+1;        sort(P + 1, P + K + 1);        ll ans = 0;        for (int i = 1; i <= K + 1; i++)            ans = (ans + C(P[i].x - P[i-1].x - 1)) % mod;        ans = ans * C(M) % mod;        for (int i = 1; i <= K; i++)            ans = (ans + solve(i)) % mod;        ans = (1LL * C(N) * C(M) % mod - ans + mod) % mod;        printf("%lld\n", ans);    }    return 0;}

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