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题目意思就是有n个干草包,将一条路分成了n-1段区间,把起点设在这n-1段的每一段中,判断从那里开始是否能够逃脱,输出所有不能逃脱的区间长度之和。

记录每个区间能不能冲出去

一个区间只要中间有一个点可以冲出去就每个点都可以冲出去

那么我们只要一直向左边和右边冲,一直到没法继续冲位置,只要冲到一个可以冲出去的区间,就可以冲出去

冲冲冲

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100011
int n;
bool b[N];
struct node{
    int v,p;
}a[N];
bool cmp(node x,node y){return x.p<y.p;}
bool chk(int p){
    int s=a[p+1].p-a[p].p,l=p,r=p+1;
    while(1<=l&&r<=n){
        bool ff=0;
        if(s>a[l].v){
            ff=1;--l;
            if(b[l])return b[p]=1,1;
            s+=a[l+1].p-a[l].p;
        }
        if(s>a[r].v){
            ff=1;++r;
            if(b[r])return b[p]=1,1;
            s+=a[r].p-a[r-1].p;
        }
        if(!ff)return 0;
    }
    return 1;
}
int main(){
    fin("in");
    in>>n;
    Fur(i,1,n)in>>a[i].v>>a[i].p;
    sort(a+1,a+n+1,cmp);
    b[0]=b[n]=1;
    int ans=0;
    Fur(i,1,n)
        if(!chk(i))ans+=a[i+1].p-a[i].p;
    out<<ans<<ln;
}
LG 3127 [USACO15OPEN]被困在haybales(金)Trappe…
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