洛谷P2070 刷墙 离散化
题意 区间覆盖 求覆盖了2次以上的线段有多少长
离散化一下，一条线段两个点，2n个点排序一下就行了
1 #include <bits/stdc++.h> 2 #define For(i, j, k) for(int i=j; i<=k; i++) 3 #define Dow(i, j, k) for(int i=j; i>=k; i--) 4 #define LL long long 5 using namespace std; 6 inline int read() { 7 int x = 0, f = 1; 8 char ch = getchar(); 9 while(ch<'0'||ch>'9') { if(ch=='-') f = -1; ch = getchar(); } 10 while(ch>='0'&&ch<='9') { x = x*10+ch-48; ch = getchar(); } 11 return x * f; 12 } 13 14 const int N = 100011; 15 struct node{ 16 int val, pos; 17 }point[2*N]; 18 int n,sum,L,R,num,ans; 19 20 inline bool cmp_pos(node a, node b) { 21 return a.pos < b.pos; 22 } 23 24 int main() { 25 n = read(); 26 int now = 0, y; 27 For(i, 1, n) { 28 char ch[10]; 29 int len = read(); scanf("%s",ch+1); 30 int x = now; 31 if(ch[1]=='L') y = x-len; else y = x+len; 32 now = y; 33 if(x>y) swap(x, y); 34 point[++num].pos = x; point[num].val = 1; 35 point[++num].pos = y; point[num].val = -1; 36 } 37 sort(point+1, point+num+1, cmp_pos); 38 39 For(i, 1, num) { 40 int old = sum; 41 sum += point[i].val; 42 if(old==1 && sum==2) L = point[i].pos; 43 if(old==2 && sum==1) ans +=point[i].pos-L; 44 } 45 printf("%d ", ans); 46 return 0; 47 }