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题意:求矩形覆盖k次以上的区域总面积。
因为k≤10,可以在线段树上维护覆盖次数为0,...,k长度数量。
然后就是一个离散化以后扫描线的问题了。
离散化用的是半开半闭区间,以方便表示没有被覆盖的区间。
/********************************************************* * --------------Alfheim-------------- * * author AbyssalFish * **********************************************************/ #include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 3e4+5; const int maxnc = maxn*2; const int maxk = 10; int x[maxnc], y[maxnc]; int xs[maxnc], mpx[maxnc]; int rx[maxnc], ry[maxnc]; int n, nxs, lim_k; #define para int o = 1, int l = 1, int r = nxs #define lo (o<<1) #define ro (o<<1|1) #define Tvar int md = (l+r)>>1; #define lsn lo,l,md #define rsn ro,md,r #define insd ql<=l&&r<=qr const int ST_SIZE = 1<<17; int sum[ST_SIZE][maxk+1]; int cnt[ST_SIZE]; #define int_byte 4 void build(para) { cnt[o] = 0; memset(sum[o]+1,0,int_byte*lim_k); sum[o][0] = mpx[r]-mpx[l]; if(r-l>1){ Tvar build(lsn); build(rsn); } } inline void maintain(para) { if(cnt[o] >= lim_k) { memset(sum[o],0,int_byte*lim_k); sum[o][lim_k] = mpx[r]-mpx[l]; } else if(r - l == 1) { int k = cnt[o]; sum[o][k] = mpx[r]-mpx[l]; if(k > 0) sum[o][k-1] = 0; if(k < lim_k) sum[o][k+1] = 0; } else { int lc = lo, rc = ro, c = cnt[o], k; for(k = 0; k < c; k++) sum[o][k] = 0; for(k = c; k <= lim_k; k++){ sum[o][k] = sum[lc][k-c] + sum[rc][k-c]; } for(k = lim_k - c+1; k <= lim_k; k++){ sum[o][lim_k] += sum[lc][k] + sum[rc][k]; } } } #define upara ql, qr, d void update(int ql, int qr, int d, para) { if(insd){ cnt[o] += d; } else { Tvar if(ql < md) update(upara,lsn); if(qr > md) update(upara,rsn); } maintain(o,l,r); } int *c_cmp; bool cmp_id(int i,int j){ return c_cmp[i] < c_cmp[j]; } bool cmp_y(int i,int j){ return y[i] < y[j] || (y[i] == y[j] && (i&1)>(j&1) ); } //出点下标i, i % 2 = 1 int compress(int n, int *a, int *r, int *b, int *mp) { for(int i = 0; i < n; i++){ r[i] = i; } c_cmp = a; sort(r,r+n,cmp_id); int k = 1; mp[b[r[0]] = 1] = a[r[0]]; for(int i = 1; i < n; i++){ int j = r[i]; if(a[j] != a[r[i-1]]){ mp[ b[j] = ++k ] = a[j]; } else { b[j] = k; } } return k; } ll solve() { scanf("%d%d",&n,&lim_k); int nn = n*2; for(int i = 0; i < nn; i++){ scanf("%d%d",x+i,y+i); ry[i] = i; } for(int i = 1; i < nn; i += 2){ //[) x[i]++; y[i]++; } nxs = compress(2*n,x,rx,xs,mpx); build(); sort(ry,ry+nn,cmp_y); ll res = 0; for(int i = 0; i < nn; i++){ int p = ry[i], q = p^1; if(i) res += (ll)sum[1][lim_k]*(y[p]-y[ry[i-1]]); if(y[p] < y[q]){ //assert((q&1) == 1); update(xs[p],xs[q],1); } else { update(xs[q],xs[p],-1); } } return res; } //#define LOCAL int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif //cout<<log2(maxnc); int T, ks = 0; scanf("%d",&T); while(++ks <= T){ printf("Case %d: %lld\n",ks,solve()); } return 0; }
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原文地址:http://www.cnblogs.com/jerryRey/p/5022050.html