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 
     
       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\n", ans); 46     return 0; 47 }