区间更新 求总区间——hdu1754

超时是因为底层结点也更新了，背离了区间更新的大原则，所以超时

View Code

#include < stdio.h >

struct data
{
int l,r,val;
}st[ 300009 ];

inline int max( int a, int b)
{
return a > b ? a:b;
}
void build( int ll, int rr, int n) // ½¨Ê÷
{
st[n].l = ll;
st[n].r = rr;
st[n].val = 1 ;
if (ll == rr) return ;
int mid = (ll + rr) / 2 ;
build(ll,mid, 2 * n);
build(mid + 1 ,rr, 2 * n + 1 );
}
void updata( int ll, int rr, int a, int n) // ¸üнáµã
{
if (st[n].l == ll && st[n].r == rr && ll == rr)
{
st[n].val = a;
return ;
}

int mid = (st[n].l + st[n].r) / 2 ;
if (rr <= mid) updata(ll,rr,a, 2 * n);
else if (ll >= mid + 1 ) updata(ll,rr,a, 2 * n + 1 );
else
{
updata(ll,mid,a, 2 * n);
updata(mid + 1 ,rr,a, 2 * n + 1 );
}
st[n].val = st[n + n].val + st[n + n + 1 ].val; // ¸üÐÂÍ·½áµã

}

int main()
{
int t;
scanf( " %d " , & t);
int add = 0 ;
while (t -- )
{
add ++ ;
int n;
scanf( " %d " , & n);

build( 1 ,n, 1 );
int tt;
scanf( " %d " , & tt);
int i;
// updata(1,n,1,1);
for (i = 1 ;i <= tt;i ++ )
{
int left,right,v;
scanf( " %d%d%d " , & left, & right, & v);
updata(left,right,v, 1 );
}
printf( " Case %d: The total value of the hook is %d.\n " ,add,st[ 1 ].val);
}
return 0 ;
}

一段区间里为纯色时，val设为非-1，为杂色是val设为-1，可以加快查找更新

View Code

#include < stdio.h >

struct data
{
int l,r,val;
}st[ 300009 ];

inline int max( int a, int b)
{
return a > b ? a:b;
}
void build( int ll, int rr, int n) // 建树
{
st[n].l = ll;
st[n].r = rr;
st[n].val = 1 ;
if (ll == rr) return ;
int mid = (ll + rr) / 2 ;
build(ll,mid, 2 * n);
build(mid + 1 ,rr, 2 * n + 1 );
}
void updata( int ll, int rr, int a, int n) // 更新结点
{
if (st[n].val == a) return ;//值同，不需要更新

if (st[n].l == ll && st[n].r == rr)
{
st[n].val = a;
return ;
}

if (st[n].val !=- 1 ) // 将纯色遗传到子代,父代变杂色
{
st[n + n].val = st[n].val;
st[n + n + 1 ].val = st[n].val;
st[n].val =- 1 ;
}

int mid = (st[n].l + st[n].r) / 2 ;
if (rr <= mid) updata(ll,rr,a, 2 * n); // 在左子树则更新
else if (ll >= mid + 1 ) updata(ll,rr,a, 2 * n + 1 ); // 在右子树则更新
else // 左右树分开时
{
updata(ll,mid,a, 2 * n);
updata(mid + 1 ,rr,a, 2 * n + 1 );
}

}

int search( int n)
{
if (st[n].val !=- 1 )
{
return st[n].val * (st[n].r - st[n].l + 1 );
}
else
{
return search(n + n) + search(n + n + 1 );
}
}

int main()
{
int t;
scanf( " %d " , & t);
int add = 0 ;
while (t -- )
{
add ++ ;
int n;
scanf( " %d " , & n);

build( 1 ,n, 1 );
int tt;
scanf( " %d " , & tt);
int i;
// updata(1,n,1,1);
for (i = 1 ;i <= tt;i ++ )
{
int left,right,v;
scanf( " %d%d%d " , & left, & right, & v);
updata(left,right,v, 1 );
}
printf( " Case %d: The total value of the hook is %d.\n " ,add,search( 1 ));
}
return 0 ;
}

还有一种枚举点，更新点的方法

View Code

#include < stdio.h >

struct data
{
int l,r,val;
}qu[ 100009 ];

int main()
{
int t,ci = 0 ;
scanf( " %d " , & t);
while (t -- )
{
ci ++ ;
int n;
scanf( " %d " , & n);

int all;
scanf( " %d " , & all);
int i,j;
for (i = 1 ;i <= all;i ++ )
{
int ll,rr,v;
scanf( " %d%d%d " , & ll, & rr, & v);
qu[i].l = ll;
qu[i].r = rr;
qu[i].val = v;
}
qu[ 0 ].val = 1 ;

int count = 0 ;
for (i = 1 ;i <= n;i ++ )
{
for (j = all;j >= 1 ;j -- )
{
if (qu[j].l <= i && i <= qu[j].r)
{
break ;
}
}
count += qu[j].val;
}
printf( " Case %d: The total value of the hook is %d.\n " ,ci,count);
}
}