敌兵布阵
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 1e5;
4 struct Node{
5 int left;
6 int right;
7 int sum;
8 } N[maxn*4];
9
10 void init(int l,int r,int num){
11 N[num].left = l;
12 N[num].right = r;
13 N[num].sum = 0;
14 if(N[num].left == N[num].right) return ;
15 int mid = (N[num].left + N[num].right) >> 1;
16 init(l, mid, num<<1);
17 init(mid+1, r, num<<1|1);
18 }
19
20 void add(int x,int y,int num){
21 if(N[num].left == N[num].right){
22 N[num].sum += y;
23 return ;
24 }
25 int mid = (N[num].left + N[num].right) >> 1;
26 if(x <= mid)
27 add(x, y, num<<1);
28 else
29 add(x, y, num<<1|1);
30 N[num].sum = N[num<<1].sum + N[num<<1|1].sum;
31 }
32
33 int findn(int ll,int rr,int num){
34 if(N[num].left == ll && N[num].right == rr){
35 return N[num].sum;
36 }
37 int mid = N[num].left + N[num].right>>1;
38 if(ll > mid){
39 return findn(ll, rr, num<<1|1);
40 }
41 else if(rr <= mid){
42 return findn(ll, rr, num<<1);
43 }
44 else
45 return findn(ll, mid, num<<1) + findn(mid+1, rr, num<<1|1);
46 }
47
48 int main(){
49 int t, n, m, countn = 1;
50 char str[20];
51 scanf("%d", &t);
52 while(t--){
53 scanf("%d", &n);
54 init(1,n,1);
55 for(int i = 1; i <= n; i++){
56 scanf("%d", &m);
57 add(i, m, 1);
58 }
59 printf("Case %d:\n", countn++);
60 while(~scanf("%s", str)){
61 if(str[0] == ‘E‘)
62 break;
63 if(str[0] == ‘A‘){
64 int a, b;
65 scanf("%d%d", &a, &b);
66 add(a, b, 1);
67 }
68 if(str[0] == ‘S‘){
69 int a, b;
70 scanf("%d%d", &a, &b);
71 add(a, -b, 1);
72 }
73 if(str[0] == ‘Q‘){
74 int a, b;
75 scanf("%d%d", &a, &b);
76 printf("%d\n", findn(a,b,1));
77 }
78 }
79 }
80 return 0;
81 }
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http://poj.org/problem?id=2828
排队。。此题妙哉
目的是求第k大数，线段树和都可以树状数组
1 #include <iostream>
2 #include <stdio.h>
3 #include <cmath>
4 using namespace std;
5 #define ls node<<1
6 #define rs node<<1|1
7 const int maxn = 200009;
8 int n, a[maxn], b[maxn], c[maxn];
9 struct Node{
10 int l, r, num;
11 }sum[maxn*4];
12 void build(int l, int r, int node){
13 sum[node].l = l, sum[node].r = r, sum[node].num = r-l+1;
14 if(l == r) return;
15 int mid = (l+r)>>1;
16 build(l, mid, ls);
17 build(mid+1, r, rs);
18 }
19 int add(int id, int node){
20 sum[node].num --;
21 if(sum[node].l == sum[node].r){
22 return sum[node].l;
23 }
24 if(sum[ls].num >= id) add(id, ls);
25 else id -= sum[ls].num, add(id, rs);
26 }
27 int main(){
28 while(~scanf("%d", &n)){
29 build(1, n, 1);
30 for(int i = 0; i < n; ++i){
31 scanf("%d%d", &a[i], &b[i]);
32 }
33 for(int i = n-1; i >= 0; --i){
34 c[add(a[i]+1, 1)] = b[i];
35 }
36 for(int i = 1; i <= n; ++i){
37 printf("%d%c", c[i], i == n ? ‘\n‘ : ‘ ‘);
38 }
39 }
40 return 0;
41 }
-------------------------------------------------------------------
http://poj.org/problem?id=3277
计算阴影面积（离散化+线段树）
1 #include <iostream>
2 #include <algorithm>
3 #include <stdio.h>
4 using namespace std;
5 #define ll long long
6 const int maxn = 204000;
7 ll l[maxn], r[maxn], h[maxn], a[maxn*2], sum[maxn*2], cnt = 0;
8 int n, num = 0, id = 1;
9 bool lz[maxn*2];
10 void p(int node){
11 if(lz[node]){
12 sum[node<<1] = max(sum[node<<1], sum[node]);
13 sum[node<<1|1] = max(sum[node<<1|1], sum[node]);
14 lz[node<<1] = lz[node<<1|1] = true;
15 lz[node] = false;
16 }
17 }
18 void add(int lx, int rx, ll val, int lxx, int rxx, int node){
19 if(lx <= lxx && rxx <= rx){
20 sum[node] = max(sum[node], val);
21 lz[node] = true;
22 return ;
23 }
24 p(node);
25 int mid = (lxx+rxx) >> 1;
26 if(lx <= mid) add(lx, rx, val, lxx, mid, node<<1);
27 if(mid < rx) add(lx, rx, val, mid+1, rxx, node<<1|1);
28 }
29 void q(int lx, int rx, int node){
30 p(node);
31 if(rx == lx){
32 cnt += (a[id]-a[id-1])*sum[node];
33 id ++;
34 return;
35 }
36 int mid = (lx+rx) >> 1;
37 q(lx, mid, node<<1);
38 q(mid+1, rx, node<<1|1);
39 }
40 int main(){
41 scanf("%d", &n);
42 for(int i = 0; i < n; ++i){
43 scanf("%lld%lld%lld", &l[i], &r[i], &h[i]);
44 a[num++] = l[i], a[num++] = r[i];
45 }
46 sort(a, a+num);
47 num = unique(a, a+num) - a;
48 for(int i = 0; i < n; ++i){
49 int xx = lower_bound(a, a+num, l[i]) - a;
50 int yy = lower_bound(a, a+num, r[i]) - a;
51 add(xx+1, yy, h[i], 1, num, 1);
52 }
53 q(1, num, 1);
54 printf("%lld\n", cnt);
55 return 0;
56 }
-------------------------------------------------------------------
http://poj.org/problem?id=2528
几张海报（离散化+线段树）
1 #include <iostream>
2 #include <string.h>
3 #include <stdio.h>
4 #include <algorithm>
5 using namespace std;
6 const int maxn = 10009;
7 struct Node{
8 int l, r;
9 }T[maxn];
10 int a[maxn*6], sum[maxn*16], cnt;
11 bool vis[maxn*6];
12 void pd(int node){
13 if(sum[node] != -1){
14 sum[node<<1] = sum[node<<1|1] = sum[node];
15 sum[node] = -1;
16 }
17 }
18 void add(int l, int r, int val, int ll, int rr, int node){
19 if(l <= ll && rr <= r){
20 sum[node] = val;
21 return;
22 }
23 pd(node);
24 int mid = (ll+rr) >> 1;
25 if(l <= mid) add(l, r, val, ll, mid, node<<1);
26 if(mid < r) add(l, r, val, mid+1, rr, node<<1|1);
27 }
28 void find(int ll, int rr, int node){
29 if(sum[node] != -1){
30 if(!vis[sum[node]]) cnt ++, vis[sum[node]] = true;
31 return;
32 }
33 if(ll == rr) return;
34 int mid = (ll+rr) >> 1;
35 find(ll, mid, node<<1);
36 find(mid+1, rr, node<<1|1);
37 }
38 int main(){
39 int t, n;
40 scanf("%d", &t);
41 while(t--){
42 int num = 0;
43 scanf("%d", &n);
44 for(int i = 0; i < n; ++i){
45 scanf("%d%d", &T[i].l, &T[i].r);
46 a[num++] = T[i].l, a[num++] = T[i].r;
47 }
48 sort(a, a+num);
49 int m = unique(a, a+num) - a;
50 int xx = m;
51 for(int i = 1; i < m; ++i){
52 if(a[i]-a[i-1] > 1) a[xx++] = a[i-1]++;
53 }
54 sort(a, a+xx);
55 memset(sum, -1, sizeof(sum));
56 for(int i = 0; i < n; ++i){
57 int l = lower_bound(a, a+xx, T[i].l) - a;
58 int r = lower_bound(a, a+xx, T[i].r) - a;
59 add(l, r, i, 0, xx, 1);
60 }
61 memset(vis, false, sizeof(vis)), cnt = 0;
62 find(0, xx, 1);
63 printf("%d\n", cnt);
64 }
65 return 0;
66 }
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只有不断学习才能进步！
原文地址：https://www.cnblogs.com/wenbao/p/5727790.html
时间： 2024-10-27 07:53:48