hdu 1698

来源：互联网发布：机器人走迷宫编程编辑：程序博客网时间：2024/06/03 18:21
线段树模板题，注意要用到lazy操作便是了。
代码如下：
#include <cstdio>#include <stack>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <list>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <iomanip>#include <cmath>#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define F first#define S second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define lrt rt << 1#define rrt rt << 1|1#define root 1,n,1#define BitCount(x) __builtin_popcount(x)#define BitCountll(x) __builtin_popcountll(x)#define LeftPos(x) 32 - __builtin_clz(x) - 1#define LeftPosll(x) 64 - __builtin_clzll(x) - 1const double PI = acos(-1.0);const LL INF = (((LL)1)<<62)+1;using namespace std;const double eps = 1e-5;const int MAXN = 300 + 10;const int MOD = 1000007;const double M=1e-8;const int N=1e6;typedef pair<int, int> pii;typedef pair<int, string> pis;const int d[4][2]={{0,1},{0,-1},{-1,0},{1,0}};int n,m,st[N],lazy[N];void pushdown(int rt,int len){    int mid=(len+1)>>1;    if (lazy[rt])    {        st[rt<<1]=mid*lazy[rt];        st[rt<<1|1]=(len-mid)*lazy[rt];        lazy[rt<<1]=lazy[rt];        lazy[rt<<1|1]=lazy[rt];        lazy[rt]=0;    }}void updata(int a,int b,int c,int l,int r,int rt){    if (a>r || b<l) return ;    if (a<=l && r<=b) {        st[rt]=c*(r-l+1);        lazy[rt]=c;        return ;    }    pushdown(rt,r-l+1);    int mid=MID(l,r);    if (a<=mid) updata(a,b,c,lson);    if (b>mid) updata(a,b,c,rson);    st[rt]=st[lrt]+st[rrt];}void build(int l,int r,int rt){    st[rt]=r-l+1;    if (l==r) {  return; }    int mid=MID(l,r);    build(lson);    build(rson);}int main(){    int i,j,T,cnt=1;    cin>>T;    while(T--)    {        scanf("%d%d",&n,&m);        build(root);        MS(lazy,0);        for (i=0;i<m;i++) {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            updata(a,b,c,root);    // for (j=1;j<n*3;j++) cout<<st[j]<<" "; cout<<endl;        }        printf("Case %d: The total value of the hook is %d.\n",cnt++,st[1]);    }}
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