【dp+离散化+线段树优化】Paint

来源：互联网发布：zbtc 数字图像算法编辑：程序博客网时间：2024/05/17 22:35
题意：求不相交的区间集合，没有覆盖到的范围最小
#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 4e5+10;const int M = 26;const int INF = 0x3f3f3f3f;vector<LL> xs;inline int getid(LL x){return lower_bound(xs.begin(),xs.end(), x) - xs.begin() + 1;}struct asd{    LL l,r;    LL sum;}a[N];bool cmp(asd x,asd y){    if(x.l != y.l) return x.l < y.l;    return x.r < y.r;}struct Node{    int left,right;    LL maxn;}node[N*4];int father[N*4];void BulitTree(int i,int l,int r){    node[i].left = l;    node[i].right = r;    node[i].maxn = 0;    if(l == r){        father[l] = i;        return ;    }    BulitTree(i*2,l,(l+r)/2);    BulitTree(i*2+1,(l+r)/2+1,r);}void Update(int i){    if(i == 1)  return ;    int pi = i/2;    LL x = node[pi*2].maxn;    LL y = node[pi*2+1].maxn;    node[pi].maxn = max(x,y);    Update(pi);}LL ans;void Query(int i,int l,int r){    if(node[i].right==r && node[i].left==l){        ans = max(ans,node[i].maxn);        return ;    }    i = i*2;    if(l <= node[i].right){        if(r <= node[i].right){            Query(i,l,r);        }        else{            Query(i,l,node[i].right);        }    }    i++;    if(r >= node[i].left){        if(l >= node[i].left)            Query(i,l,r);        else            Query(i,node[i].left,r);    }}int main(){    LL n;    int m;    scanf("%lld%d",&n,&m);    BulitTree(1,1,m*2+1);    for(int i = 1; i <= m; i++){        scanf("%I64d%I64d",&a[i].l,&a[i].r);        a[i].sum = a[i].r-a[i].l+1;    }    sort(a+1,a+m+1,cmp);    xs.push_back(0);    for(int i = 1; i <= m; i++)    {        xs.push_back(a[i].l);        xs.push_back(a[i].r);    }    sort(xs.begin(), xs.end());    vector<LL>::iterator e;    e=unique(xs.begin(),xs.end());    xs.erase(unique(xs.begin(), xs.end()) , xs.end());    int id = getid(0);    int x = father[id];    node[x].maxn = 0;    Update(x);    for(int i = 1; i <= m; i++){        id = getid(a[i].l);        ans = 0;        Query(1,1,id-1);        id = getid(a[i].r);        x = father[id];        node[x].maxn = ans+a[i].sum;        Update(x);    }    ans = 0;    Query(1,1,m*2+1);    printf("%I64d\n",n-ans);    return 0;}
阅读全文
0 0