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Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23644 Accepted Submission(s): 11839
#include <cstdio> #include <iostream> #include <cstdlib> #include <algorithm> #include <ctime> #include <cmath> #include <string> #include <cstring> #include <stack> #include <queue> #include <list> #include <vector> #include <map> #include <set> using namespace std; const int INF=0x3f3f3f3f; const double eps=1e-10; const double PI=acos(-1.0); #define maxn 110000 int tre[maxn*4]; int lazy[maxn*4]; void build(int num, int le, int ri) { if(le == ri) { tre[num] = 1; return; } int mid = (le+ri)/2; build(num*2, le, mid); build(num*2+1, mid+1, ri); tre[num] = tre[num*2] + tre[num*2+1]; } void pushdown(int num, int le, int mid, int ri) { if(lazy[num]!=0) { lazy[num*2] = lazy[num]; lazy[num*2+1] = lazy[num]; tre[num*2] = (mid-le+1)*lazy[num*2]; tre[num*2+1] = (ri-mid)*lazy[num*2+1]; lazy[num] = 0; } } void update(int num, int le, int ri, int x, int y, int val) { if(x<=le && y>=ri) { tre[num] = (ri-le+1)*val; lazy[num] = val; return; } int mid = (le+ri)/2; pushdown(num,le,mid,ri); if(x<=mid) update(num*2,le,mid,x,y,val); if(y>mid) update(num*2+1,mid+1,ri,x,y,val); tre[num] = tre[num*2] + tre[num*2+1]; } int main() { int t; scanf("%d", &t); int cas = 0; while(t--) { int n; scanf("%d", &n); build(1,1,n); memset(lazy, 0, sizeof lazy); int m; scanf("%d", &m); for(int i = 0; i < m; i++) { int a,b,c; scanf("%d%d%d", &a, &b, &c); update(1,1,n,a,b,c); } printf("Case %d: The total value of the hook is %d.\n", ++cas, tre[1]); } return 0; }
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原文地址:http://www.cnblogs.com/ZP-Better/p/4857846.html