hdu 4366 Successor
做法：对每个人按照ability由大到小排序，把loyalty插入到线段树里面，dfs处理出每个点所在的区间，然后区间查询，单点更新。（这里学到了查询区间最大值所在id的方法）。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <queue> 7 #include <stack> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-6 13 #define inf 0x3f3f3f3f 14 #define MP make_pair 15 #define N 250020 16 #define M 1000020 17 #pragma comment(linker, "/STACK:1024000000,1024000000") 18 #define Pi acos(-1.0) 19 #define ls (i << 1) 20 #define rs (ls | 1) 21 #define md ((ll + rr) >> 1) 22 #define lson ll, md, ls 23 #define rson md + 1, rr, rs 24 25 int fst[N], vv[N], nxt[N], e; 26 void init(){ 27 memset(fst, -1, sizeof fst); e = 0; 28 } 29 void add(int u, int v){ 30 vv[e] = v, nxt[e] = fst[u], fst[u] = e++; 31 } 32 int in[N], out[N], lab[N], dc; 33 void dfs(int u){ 34 in[u] = ++dc; 35 lab[dc] = u; 36 for(int i = fst[u]; ~i; i = nxt[i]){ 37 int v = vv[i]; 38 dfs(v); 39 } 40 out[u] = dc; 41 } 42 struct node{ 43 int id, val, loy; 44 node(){} 45 node(int _loy, int _val, int _id){ 46 loy = _loy, val = _val, id = _id; 47 } 48 bool operator < (const node &b) const{ 49 return val > b.val; 50 } 51 }b[N]; 52 53 int mx[N<<2], val[N]; 54 void build(int ll, int rr, int i){ 55 mx[i] = -1; 56 if(ll == rr) return ; 57 build(lson), build(rson); 58 } 59 void push_up(int i){ 60 if(val[mx[ls]] > val[mx[rs]]) 61 mx[i] = mx[ls]; 62 else mx[i] = mx[rs]; 63 } 64 void update(int p, int v, int ll, int rr, int i){ 65 if(ll == rr){ 66 val[ll] = v; 67 mx[i] = ll; 68 return ; 69 } 70 if(p <= md) update(p, v, lson); 71 else update(p, v, rson); 72 push_up(i); 73 } 74 int query(int l, int r, int ll, int rr, int i){ 75 if(l > r) return -1; 76 if(ll == l && r == rr) 77 return mx[i]; 78 if(r <= md) return query(l, r, lson); 79 else if(l > md) return query(l, r, rson); 80 else{ 81 int ret1 = query(l, md, lson); 82 int ret2 = query(md + 1, r, rson); 83 if(ret1 == -1) return ret2; 84 if(ret2 == -1) return ret1; 85 return val[ret1] >= val[ret2] ? ret1 : ret2; 86 } 87 } 88 89 int ans[N]; 90 int main(){ 91 int cas; 92 scanf("%d", &cas); 93 while(cas--){ 94 int n, m; 95 scanf("%d%d", &n, &m); 96 init(); 97 for(int i = 1; i < n; ++i){ 98 int u, val, loy; 99 scanf("%d%d%d", &u, &loy, &val); 100 add(u, i); 101 b[i] = node(loy, val, i); 102 } 103 dc = 0; 104 dfs(0); 105 sort(b + 1, b + n); 106 build(1, n, 1); 107 memset(val, -1, sizeof val); 108 for(int i = 1, j; i < n; i = j){ 109 j = i; 110 while(j < n && b[i].val == b[j].val){ 111 int k = b[j].id; 112 int t = query(in[k] + 1, out[k], 1, n, 1); 113 if(t == -1) ans[k] = -1; 114 else ans[k] = lab[t]; 115 j++; 116 } 117 j = i; 118 while(j < n && b[i].val == b[j].val){ 119 update(in[b[j].id], b[j].loy, 1, n, 1); 120 j++; 121 } 122 } 123 while(m--){ 124 int i; 125 scanf("%d", &i); 126 printf("%d ", ans[i]); 127 } 128 } 129 return 0; 130 }
hdu 3397 Sequence operation
区间赋值，异或。记录左边连续最大，右边连续最大，总的连续最大，区间1和0的个数。（询问的时候返回连续最大的1的个数那里老是写挫了！！！）
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <queue> 7 using namespace std; 8 9 #define LL long long 10 #define eps 1e-8 11 #define inf 0x3f3f3f3f 12 #define MP make_pair 13 #define N 100020 14 #define M 200020 15 #pragma comment(linker, "/STACK:1024000000,1024000000") 16 #define Pi acos(-1.0) 17 #define mod 1000000007 18 #define ls (i << 1) 19 #define rs (ls | 1) 20 #define md ((ll + rr) >> 1) 21 #define lson ll, md, ls 22 #define rson md + 1, rr, rs 23 24 int mxL[2][N<<2], mxR[2][N<<2], mx[2][N<<2]; 25 int sum[2][N<<2], cov[N<<2], Xor[N<<2]; 26 27 void init(int x, int i, int len){ 28 mxL[x][i] = mxR[x][i] = mx[x][i] = len; 29 sum[x][i] = len; 30 } 31 32 void push_up(int x, int i, int ll, int rr){ 33 mxL[x][i] = mxL[x][ls], mxR[x][i] = mxR[x][rs]; 34 mx[x][i] = max(mx[x][ls], mx[x][rs]); 35 mx[x][i] = max(mx[x][i], mxR[x][ls] + mxL[x][rs]); 36 if(mxL[x][ls] == md - ll + 1) 37 mxL[x][i] += mxL[x][rs]; 38 if(mxR[x][rs] == rr - md) 39 mxR[x][i] += mxR[x][ls]; 40 sum[x][i] = sum[x][ls] + sum[x][rs]; 41 } 42 void build(int ll, int rr, int i){ 43 cov[i] = -1, Xor[i] = 0; 44 if(ll == rr){ 45 int v; 46 scanf("%d", &v); 47 init(v, i, 1); 48 init(v^1, i, 0); 49 return ; 50 } 51 build(lson), build(rson); 52 push_up(0, i, ll, rr); 53 push_up(1, i, ll, rr); 54 } 55 void change(int i){ 56 Xor[i] ^= 1; 57 swap(mxL[0][i], mxL[1][i]); 58 swap(mxR[0][i], mxR[1][i]); 59 swap(mx[0][i], mx[1][i]); 60 swap(sum[0][i], sum[1][i]); 61 } 62 void push_down(int i, int ll, int rr){ 63 if(cov[i] != -1){ 64 int x = cov[i]; 65 cov[ls] = cov[rs] = x; 66 Xor[ls] = Xor[rs] = 0; 67 init(x, ls, md - ll + 1); 68 init(x, rs, rr - md); 69 init(x^1, ls, 0); 70 init(x^1, rs, 0); 71 cov[i] = -1; 72 } 73 if(Xor[i]){ 74 change(ls), change(rs); 75 Xor[i] = 0; 76 } 77 } 78 void update(int l, int r, int op, int ll, int rr, int i){ 79 if(l == ll && r == rr){ 80 if(op == 0 || op == 1){ 81 init(op, i, rr - ll + 1); 82 init(op ^ 1, i, 0); 83 cov[i] = op, Xor[i] = 0; 84 } 85 else{ 86 change(i); 87 } 88 return ; 89 } 90 push_down(i, ll, rr); 91 if(r <= md) update(l, r, op, lson); 92 else if(l > md) update(l, r, op, rson); 93 else 94 update(l, md, op, lson), update(md + 1, r, op, rson); 95 push_up(0, i, ll, rr); 96 push_up(1, i, ll, rr); 97 } 98 99 int query(int l, int r, int op, int ll, int rr, int i){ 100 if(l == ll && r == rr){ 101 if(op == 4) 102 return mx[1][i]; 103 else return sum[1][i]; 104 } 105 push_down(i, ll, rr); 106 int ret; 107 if(r <= md) ret = query(l, r, op, lson); 108 else if(l > md) ret = query(l, r, op, rson); 109 else{ 110 int ret1 = query(l, md, op, lson); 111 int ret2 = query(md + 1, r, op, rson); 112 if(op == 4){ 113 ret = max(ret1, ret2); 114 ret = max(ret, min(mxR[1][ls], md - l + 1) + min(mxL[1][rs], r - md)); 115 } 116 else ret = ret1 + ret2; 117 } 118 push_up(0, i, ll, rr); 119 push_up(1, i, ll, rr); 120 return ret; 121 } 122 int main(){ 123 int cas; 124 //freopen("tt.txt", "r", stdin); 125 scanf("%d", &cas); 126 while(cas--){ 127 int n, m; 128 scanf("%d%d", &n, &m); 129 build(1, n, 1); 130 while(m--){ 131 int op, L, R; 132 scanf("%d%d%d", &op, &L, &R); 133 ++L, ++R; 134 if(op <= 2) 135 update(L, R, op, 1, n, 1); 136 else printf("%d ", query(L, R, op, 1, n, 1)); 137 } 138 } 139 return 0; 140 }
hdu 2871 Memory Control
区间更新，合并。。好题。。以前敲的时候感觉好难，现在就感觉还行。（细节没注意。push_down的时候tot[i] == 0的时候没写，wa了一发。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <queue> 7 using namespace std; 8 9 #define LL long long 10 #define eps 1e-8 11 #define inf 0x3f3f3f3f 12 #define MP make_pair 13 #define N 50020 14 #define M 200020 15 #pragma comment(linker, "/STACK:1024000000,1024000000") 16 #define Pi acos(-1.0) 17 #define mod 1000000007 18 #define ls (i << 1) 19 #define rs (ls | 1) 20 #define md ((ll + rr) >> 1) 21 #define lson ll, md, ls 22 #define rson md + 1, rr, rs 23 24 int cov[N<<2], mxL[N<<2], mxR[N<<2], mx[N<<2], tot[N<<2], id[N<<2], L[N], R[N]; 25 void build(int ll, int rr, int i){ 26 cov[i] = -1, id[i] = 0, tot[i] = 0; 27 mxL[i] = mxR[i] = mx[i] = rr - ll + 1; 28 if(ll == rr) return ; 29 build(lson), build(rson); 30 } 31 32 int cnt; 33 void f(int i, int len){ 34 mxL[i] = mxR[i] = mx[i] = len; 35 } 36 void push_down(int i, int ll, int rr){ 37 if(cov[i] != -1){ 38 cov[ls] = cov[rs] = cov[i]; 39 id[ls] = id[rs] = id[i]; 40 if(cov[i] == 0){ 41 f(ls, md - ll + 1); 42 f(rs, rr - md); 43 } 44 else{ 45 f(ls, 0); 46 f(rs, 0); 47 } 48 if(tot[i] == 0) tot[ls] = tot[rs] = 0; 49 cov[i] = -1; 50 } 51 } 52 void push_up(int i, int ll, int rr){ 53 mxL[i] = mxL[ls], mxR[i] = mxR[rs]; 54 mx[i] = max(mxR[ls] + mxL[rs], max(mx[ls], mx[rs])); 55 tot[i] = tot[ls] + tot[rs]; 56 if(mxL[ls] == md - ll + 1) 57 mxL[i] += mxL[rs]; 58 if(mxR[rs] == rr - md) 59 mxR[i] += mxR[ls]; 60 } 61 void update(int l, int r, int v, int ll, int rr, int i){ 62 if(l == ll && r == rr){ 63 cov[i] = v; 64 if(v == 0) 65 id[i] = 0, mxL[i] = mxR[i] = mx[i] = rr - ll + 1; 66 else 67 id[i] = cnt, mxL[i] = mxR[i] = mx[i] = 0; 68 return ; 69 } 70 push_down(i, ll, rr); 71 if(r <= md) update(l, r, v, lson); 72 else if(l > md) update(l, r, v, rson); 73 else 74 update(l, md, v, lson), update(md + 1, r, v, rson); 75 push_up(i, ll, rr); 76 } 77 void update(int p, int v, int ll, int rr, int i){ 78 if(ll == rr){ 79 tot[i] = v; return ; 80 } 81 push_down(i, ll, rr); 82 if(p <= md) update(p, v, lson); 83 else update(p, v, rson); 84 push_up(i, ll, rr); 85 } 86 int query(int v, int ll, int rr, int i){ 87 if(ll == rr) 88 return ll; 89 push_down(i, ll, rr); 90 int ret; 91 if(mx[ls] >= v) 92 ret = query(v, lson); 93 else if(mxR[ls] + mxL[rs] >= v) 94 ret = md - mxR[ls] + 1; 95 else 96 ret = query(v, rson); 97 push_up(i, ll, rr); 98 return ret; 99 } 100 int find(int p, int ll, int rr, int i){ 101 if(ll == rr) 102 return id[i]; 103 push_down(i, ll, rr); 104 int ret; 105 if(p <= md) ret = find(p, lson); 106 else ret = find(p, rson); 107 push_up(i, ll, rr); 108 return ret; 109 } 110 int get(int v, int ll, int rr, int i){ 111 if(ll == rr) 112 return ll; 113 push_down(i, ll, rr); 114 int ret; 115 if(tot[ls] >= v) ret = get(v, lson); 116 else ret = get(v - tot[ls], rson); 117 push_up(i, ll, rr); 118 return ret; 119 } 120 int main(){ 121 int n, m; 122 while(scanf("%d%d", &n, &m) != EOF){ 123 build(1, n, 1); 124 char s[10]; 125 while(m--){ 126 scanf("%s", s); 127 if(s[0] == 'R'){ 128 update(1, n, 0, 1, n, 1); 129 cnt = 0; 130 tot[1] = 0; 131 puts("Reset Now"); 132 continue; 133 } 134 int v; scanf("%d", &v); 135 if(s[0] == 'N'){ 136 if(mx[1] < v){ 137 puts("Reject New"); continue; 138 } 139 int ans = query(v, 1, n, 1); 140 printf("New at %d ", ans); 141 L[++cnt] = ans, R[cnt] = ans + v - 1; 142 update(ans, ans + v - 1, 1, 1, n, 1); 143 update(ans, 1, 1, n, 1); 144 } 145 else if(s[0] == 'F'){ 146 int ret = find(v, 1, n, 1); 147 if(ret == 0){ 148 puts("Reject Free"); continue; 149 } 150 printf("Free from %d to %d ", L[ret], R[ret]); 151 update(L[ret], R[ret], 0, 1, n, 1); 152 update(L[ret], 0, 1, n, 1); 153 } 154 else{ 155 if(tot[1] < v){ 156 puts("Reject Get"); continue; 157 } 158 printf("Get at %d ", get(v, 1, n, 1)); 159 } 160 } 161 puts(""); 162 } 163 return 0; 164 }
hdu 1540 Tunnel Warfare
区间合并。。没什么难度了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <queue> 7 using namespace std; 8 9 #define LL long long 10 #define eps 1e-8 11 #define inf 0x3f3f3f3f 12 #define MP make_pair 13 #define N 100020 14 #define M 200020 15 #pragma comment(linker, "/STACK:1024000000,1024000000") 16 #define Pi acos(-1.0) 17 #define mod 1000000007 18 #define ls (i << 1) 19 #define rs (ls | 1) 20 #define md ((ll + rr) >> 1) 21 #define lson ll, md, ls 22 #define rson md + 1, rr, rs 23 24 int mxL[N<<2], mxR[N<<2], mx[N<<2]; 25 int a[N], tot; 26 bool vis[N]; 27 void build(int ll, int rr, int i){ 28 mxL[i] = mxR[i] = mx[i] = rr - ll + 1; 29 if(ll == rr) return; 30 build(lson), build(rson); 31 } 32 33 void push_up(int i, int ll, int rr){ 34 mxL[i] = mxL[ls], mxR[i] = mxR[rs]; 35 mx[i] = max(mxR[ls] + mxL[rs], max(mx[ls], mx[rs])); 36 if(mxL[i] == md - ll + 1) 37 mxL[i] += mxL[rs]; 38 if(mxR[i] == rr - md) 39 mxR[i] += mxR[ls]; 40 } 41 void update(int p, int v, int ll, int rr, int i){ 42 if(ll == rr){ 43 mxL[i] = mxR[i] = mx[i] = v; 44 return ; 45 } 46 if(p <= md) update(p, v, lson); 47 else update(p, v, rson); 48 push_up(i, ll, rr); 49 } 50 51 int queryL(int l, int r, int ll, int rr, int i){ 52 if(l == ll && r == rr) 53 return mxR[i]; 54 if(r <= md) return queryL(l, r, lson); 55 if(l > md) return queryL(l, r, rson); 56 int ret = queryL(md + 1, r, rson); 57 if(ret == r - md) 58 ret += queryL(l, md, lson); 59 return ret; 60 } 61 int queryR(int l, int r, int ll, int rr, int i){ 62 if(l == ll && r == rr) 63 return mxL[i]; 64 if(r <= md) return queryR(l, r, lson); 65 if(l > md) return queryR(l, r, rson); 66 int ret = queryR(l, md, lson); 67 if(ret == md - l + 1) 68 ret += queryR(md + 1, r, rson); 69 return ret; 70 } 71 int main(){ 72 int n, m; 73 // freopen("tt.txt", "r", stdin); 74 while(scanf("%d%d", &n, &m) != EOF){ 75 build(1, n, 1); 76 memset(vis, 0, sizeof vis); 77 tot = 0; 78 char s[2]; 79 int p; 80 while(m--){ 81 scanf("%s", s); 82 if(s[0] == 'D'){ 83 scanf("%d", &p); 84 a[++tot] = p; 85 vis[p] = 1; 86 update(p, 0, 1, n, 1); 87 } 88 else if(s[0] == 'Q'){ 89 scanf("%d", &p); 90 if(vis[p]){ 91 puts("0"); continue; 92 } 93 int ret = 1; 94 if(p - 1 > 0) 95 ret += queryL(1, p - 1, 1, n, 1); 96 if(p + 1 <= n) 97 ret += queryR(p + 1, n, 1, n, 1); 98 printf("%d ", ret); 99 } 100 else{ 101 if(tot > 0){ 102 vis[a[tot]] = 0; 103 update(a[tot--], 1, 1, n, 1); 104 } 105 } 106 } 107 } 108 return 0; 109 }
hdu 4288 Coder
离散化，单点更新，要用long long。num[i]表示区间i内有num[i]个存在的元素。然后sum[j][i] = sum[j][ls] + sum[(5-num[ls]%5+j)%5]（一开始推错了）。。最后询问的时候输出sum[3][1]就好了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <queue> 7 using namespace std; 8 9 #define LL long long 10 #define eps 1e-8 11 #define inf 0x3f3f3f3f 12 #define MP make_pair 13 #define N 100020 14 #define M 200020 15 #pragma comment(linker, "/STACK:1024000000,1024000000") 16 #define Pi acos(-1.0) 17 #define mod 1000000007 18 #define ls (i << 1) 19 #define rs (ls | 1) 20 #define md ((ll + rr) >> 1) 21 #define lson ll, md, ls 22 #define rson md + 1, rr, rs 23 24 LL sum[5][N<<2]; 25 int num[N<<2]; 26 void push_up(int i){ 27 num[i] = num[ls] + num[rs]; 28 int tmp = 5 - num[ls] % 5; 29 for(int j = 0; j < 5; ++j) 30 sum[j][i] = sum[j][ls] + sum[(j+tmp) % 5][rs]; 31 } 32 33 int a[N], b[N], op[N]; 34 void update(int p, int op, int ll, int rr, int i){ 35 if(ll == rr){ 36 sum[1][i] = op ? b[p] : 0; 37 num[i] = op; 38 return ; 39 } 40 if(p <= md) update(p, op, lson); 41 else update(p, op, rson); 42 push_up(i); 43 } 44 45 int main(){ 46 //freopen("tt.txt", "r", stdin); 47 int m; 48 while(scanf("%d", &m) != EOF){ 49 memset(sum, 0, sizeof sum); 50 memset(num, 0, sizeof num); 51 int n = 0, tot = 0, v; 52 char s[5]; 53 for(int i = 0; i < m; ++i){ 54 scanf("%s", s); 55 if(s[0] == 'a'){ 56 scanf("%d", &a[i]); 57 op[i] = 1; 58 b[++n] = a[i]; 59 } 60 else if(s[0] == 'd'){ 61 scanf("%d", &a[i]); 62 op[i] = 0; 63 } 64 else op[i] = 2; 65 } 66 sort(b + 1, b + n + 1); 67 for(int i = 0; i < m; ++i){ 68 if(op[i] == 1 || op[i] == 0){ 69 int p = lower_bound(b + 1, b + n + 1, a[i]) - b; 70 update(p, op[i], 1, n, 1); 71 } 72 else printf("%lld ", sum[3][1]); 73 } 74 } 75 return 0; 76 }
hdu 4521 小明系列问题——小明序列
给你一个序列，求最长上升子序列且 序列中相邻的数的下标要大于d。。
按照 x 排序，x相同就按照下标由大到小排。按 x值 扫过去，用线段树维护下标id的最长上升序列的值。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<string> 5 #include<cstring> 6 #include<algorithm> 7 #include<set> 8 using namespace std; 9 10 #define ULL unsigned long long 11 #define eps 1e-9 12 #define inf 0x3f3f3f3f 13 #define FOR(i,s,t) for(int i = s; i < t; ++i ) 14 #define REP(i,s,t) for( int i = s; i <= t; ++i ) 15 #define pii pair<int,int> 16 #define MP make_pair 17 #define ls i << 1 18 #define rs ls | 1 19 #define md ((ll + rr) >> 1) 20 #define lson ll, md, ls 21 #define rson md + 1, rr, rs 22 #define LL long long 23 #define N 150010 24 #define M 200020 25 #define mod 1000000007 26 27 struct node{ 28 int x, id; 29 bool operator < (const node &b) const{ 30 return x < b.x || (x == b.x && id > b.id); 31 } 32 }b[N]; 33 int mx[N<<2]; 34 void update(int p, int v, int ll, int rr, int i){ 35 if(ll == rr){ 36 mx[i] = max(mx[i], v); 37 return ; 38 } 39 if(p <= md) update(p, v, lson); 40 else update(p, v, rson); 41 mx[i] = max(mx[ls], mx[rs]); 42 } 43 int query(int l, int r, int ll, int rr, int i){ 44 if(l == ll && r == rr) 45 return mx[i]; 46 if(r <= md) return query(l, r, lson); 47 if(l > md) return query(l, r, rson); 48 return max(query(l, md, lson), query(md + 1, r, rson)); 49 } 50 int a[N]; 51 int main(){ 52 int n, d; 53 while(scanf("%d%d", &n, &d) != EOF){ 54 for(int i = 1; i <= n; ++i){ 55 scanf("%d", &a[i]); 56 b[i].x = a[i], b[i].id = i; 57 } 58 sort(b + 1, b + 1 + n); 59 int ans = 0, dp; 60 memset(mx, 0, sizeof mx); 61 for(int i = 1; i <= n; ++i){ 62 if(b[i].id - d - 1 > 0) 63 dp = query(1, b[i].id - d - 1, 1, n, 1) + 1; 64 else dp = 1; 65 ans = max(dp, ans); 66 update(b[i].id, dp, 1, n, 1); 67 } 68 printf("%d ", ans); 69 } 70 return 0; 71 }
hdu 1542 Atlantis
矩形面积并。。发现不用push_down，感觉好厉害。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<string> 5 #include<cstring> 6 #include<algorithm> 7 #include<set> 8 using namespace std; 9 10 #define ULL unsigned long long 11 #define eps 1e-9 12 #define inf 0x3f3f3f3f 13 #define ls i << 1 14 #define rs ls | 1 15 #define md ((ll + rr) >> 1) 16 #define lson ll, md, ls 17 #define rson md + 1, rr, rs 18 #define LL long long 19 #define N 150010 20 #define M 200020 21 #define mod 1000000007 22 23 struct edge{ 24 double ux, vx, y, in; 25 edge(double ux = 0, double vx = 0, double y = 0, int in = 0) : ux(ux), vx(vx), y(y), in(in) {} 26 bool operator < (const edge &b) const{ 27 return y < b.y; 28 } 29 }e[N]; 30 31 double sum[N<<2], x[N]; 32 int vis[N<<2]; 33 void build(int ll, int rr, int i){ 34 vis[i] = 0, sum[i] = 0; 35 if(ll == rr) return ; 36 build(lson), build(rson); 37 } 38 void push_up(int i, int ll, int rr){ 39 if(vis[i]) sum[i] = x[rr+1] - x[ll]; 40 else if(ll == rr) sum[i] = 0; 41 else sum[i] = sum[ls] + sum[rs]; 42 } 43 void update(int l, int r, int v, int ll, int rr, int i){ 44 if(l == ll && r == rr){ 45 vis[i] += v; 46 push_up(i, ll, rr); 47 return ; 48 } 49 if(r <= md) update(l, r, v, lson); 50 else if(l > md) update(l, r, v, rson); 51 else update(l, md, v, lson), update(md + 1, r, v, rson); 52 push_up(i, ll, rr); 53 } 54 int main(){ 55 int n, kk = 0; 56 while(scanf("%d", &n) != EOF && n){ 57 printf("Test case #%d ", ++kk); 58 for(int i = 0; i < n; ++i){ 59 double ax, ay, bx, by; 60 scanf("%lf%lf%lf%lf", &ax, &ay, &bx, &by); 61 x[i] = ax, e[i] = edge(ax, bx, ay, 1); 62 x[i+n] = bx, e[i+n] = edge(ax, bx, by, -1); 63 } 64 n <<= 1; 65 sort(x, x + n); 66 sort(e, e + n); 67 int k = 1; 68 for(int i = 1; i < n; ++i) 69 if(x[i] != x[i-1]) x[k++] = x[i]; 70 build(0, k, 1); 71 double ans = 0; 72 for(int i = 0; i < n - 1; ++i){ 73 int l = lower_bound(x, x + k, e[i].ux) - x; 74 int r = lower_bound(x, x + k, e[i].vx) - x; 75 if(l < r) 76 update(l, r - 1, e[i].in, 0, k, 1); 77 ans += sum[1] * (e[i+1].y - e[i].y); 78 } 79 printf("Total explored area: %.2lf ", ans); 80 puts(""); 81 } 82 return 0; 83 }