UVA 11261 - Bishops（杂题）

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题目链接：点击打开链接

思路：

象可以沿着对角线走任意距离，直接枚举保存复杂度n*m*log(nm)，肯定超时，考虑到数学方法：同主对角线上y-x的值相同，同一副对角线上x+y相同，且连续分布。

所以我们考虑枚举主对角线的值，如果这个值出现过，那么这一个对角线的所有点都会被攻击，否则，我们维护这个区间的副对角线被占有的值的个数。

复杂度O(nlogn)

细节参见代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <ctime>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const double PI = acos(-1);const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;// & 0x7FFFFFFFconst int seed = 131;const ll INF64 = ll(1e18);const int maxn = 4e4 + 10;int T,n,m,vis[maxn*2], sum1[maxn*2], sum2[maxn*2], kase = 0;struct node {    int x, y;    node(int x=0, int y=0):x(x), y(y) {}    bool operator < (const node& rhs) const {        if(x != rhs.x) return x < rhs.x;        else return y < rhs.y;    }}a[maxn];map<int, int> p;int main() {    scanf("%d",&T);    while(T--) {        scanf("%d%d", &n, &m);        p.clear();        memset(vis, 0, sizeof(vis));        for(int i = 1; i <= m; i++) {            scanf("%d%d", &a[i].x, &a[i].y);            a[i].x--; a[i].y--;            int cur = a[i].x + a[i].y;            vis[cur] = 1;            cur = a[i].y - a[i].x;            p[cur] = 1;        }        for(int i = 0; i <= 2*(n-1); i++) {            if(i) {                if(i&1) sum1[i] = sum1[i-1] + vis[i], sum2[i] = sum2[i-1];                else sum2[i] = sum2[i-1] + vis[i], sum1[i] = sum1[i-1];            }            else {                if(i&1) sum1[i] = vis[i], sum2[i] = 0;                else sum2[i] = vis[i], sum1[i] = 0;            }        }        int l = n-1, r = n-1, ans = 0;        for(int i = -(n-1); i <= (n-1); i++) {            if(p.count(i)) {                ans += (r - l)/2+1;            }            else {                if(r & 1) ans += sum1[r] - sum1[l-1];                else ans += sum2[r] - sum2[l-1];            }            if(i < 0) l--, r++;            else l++, r--;        }        printf("Case #%d: %d\n", ++kase, n*n-ans);    }    return 0;}

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