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1 /************************************************************** 2 Problem: 1911 3 User: mjy0724 4 Language: C++ 5 Result: Accepted 6 Time:1688 ms 7 Memory:24248 kb 8 ****************************************************************/ 9 10 #include<cstdio> 11 #include<cstdlib> 12 #include<cstring> 13 const int maxn=1000010;long long INF=10000000000000007; 14 long long n,a,b,c,sum[maxn],opt[maxn],f[maxn]; 15 long long g(int j,int k){ 16 if (2*a*(sum[j]-sum[k])==0) 17 { 18 if (f[j]+a*sum[j]*sum[j]-b*sum[j]-(f[k]+a*sum[k]*sum[k]-b*sum[k])>0) return INF; 19 else return -INF; 20 } 21 return (f[j]+a*sum[j]*sum[j]-b*sum[j]-(f[k]+a*sum[k]*sum[k]-b*sum[k]))/(2*a*(sum[j]-sum[k])); 22 } 23 24 int main(){ 25 scanf("%lld",&n); 26 scanf("%lld%lld%lld",&a,&b,&c); 27 for (int i=2;i<=n+1;i++) scanf("%lld",&sum[i]),sum[i]+=sum[i-1]; 28 int head=1,tail=1;opt[1]=1;f[1]=0; 29 for (int i=2;i<=n+1;i++){ 30 if (a>=0) while (head<tail&&g(opt[head],opt[head+1])>sum[i]) head++; 31 else while (head<tail&&g(opt[head],opt[head+1])<sum[i]) head++; 32 int j=opt[head]; 33 f[i]=f[j]+a*(sum[i]-sum[j])*(sum[i]-sum[j])+b*(sum[i]-sum[j])+c; 34 if (a>=0) while (head<tail&&g(opt[tail-1],opt[tail])<g(opt[tail],i)) tail--; 35 else while (head<tail&&g(opt[tail-1],opt[tail])>g(opt[tail],i)) tail--; 36 opt[++tail]=i; 37 } 38 printf("%lld\n",f[n+1]); 39 return 0; 40 }
1 /************************************************************** 2 Problem: 1912 3 User: mjy0724 4 Language: C++ 5 Result: Accepted 6 Time:708 ms 7 Memory:5288 kb 8 ****************************************************************/ 9 10 #include<cstdio> 11 #include<cstdlib> 12 #include<cstring> 13 const int maxn=100010,maxm=200010; 14 int n,k,e,link[maxn],next[maxm],fa[maxm],pos1[maxn],pos2[maxn],w[maxm],ans,sum,s; 15 void add(int x,int y){ 16 fa[++e]=y;next[e]=link[x];link[x]=e;w[e]=1; 17 fa[++e]=x;next[e]=link[y];link[y]=e;w[e]=1; 18 } 19 20 int dfs(int p,int fat){ 21 int max1=0,max2=0; 22 for (int i=link[p];i;i=next[i]) if (fa[i]!=fat){ 23 int tmp=dfs(fa[i],p)+w[i]; 24 if (tmp>max1){ 25 max2=max1;max1=tmp;pos2[p]=pos1[p];pos1[p]=i; 26 }else if (tmp>max2) max2=tmp,pos2[p]=i; 27 } 28 if (max1+max2>ans) { 29 ans=max1+max2;s=p; 30 } 31 return max1; 32 } 33 34 int main(){ 35 scanf("%d%d",&n,&k); 36 int e=0,x,y; 37 for (int i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y); 38 dfs(1,0);sum=ans; 39 if (k==1){printf("%d\n",2*(n-1)-(ans-1));return 0;} 40 for (int i=pos1[s];i;i=pos1[fa[i]]) w[i]=-1; 41 for (int i=pos2[s];i;i=pos1[fa[i]]) w[i]=-1; 42 ans=0;dfs(1,0); 43 printf("%d\n",2*(n-1)-(ans-1)-(sum-1)); 44 return 0; 45 }
1 /************************************************************** 2 Problem: 1913 3 User: mjy0724 4 Language: C++ 5 Result: Accepted 6 Time:1996 ms 7 Memory:1356 kb 8 ****************************************************************/ 9 10 #include<cstdio> 11 #include<cstdlib> 12 #include<cstring> 13 #include<cmath> 14 #include<algorithm> 15 #include<iostream> 16 #define ll long long 17 #define maxn 1510 18 struct point{ 19 ll x,y; 20 double angle; 21 }; 22 ll n,an,b; 23 point p[maxn],a[maxn]; 24 ll cross(point p0,point p1,point p2){ 25 return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 26 } 27 void sortf(int l,int r){ 28 int i=l,j=r;double mid=p[(l+r) >> 1].angle; 29 do{ 30 while (i<r&&p[i].angle<mid) i++; 31 while (l<j&&p[j].angle>mid) j--; 32 if (i<=j){ 33 point tmp=p[i];p[i]=p[j];p[j]=tmp; 34 i++;j--; 35 } 36 }while (i<=j); 37 if (i<r) sortf(i,r); 38 if (l<j) sortf(l,j); 39 } 40 ll calc(int x){ 41 int tot=0; 42 for (int i=1;i<=n;i++) {if (i!=x) p[++tot]=a[i];else p[0]=a[i];} 43 for (int i=1;i<=tot;i++) p[i].angle=atan2(p[i].y-p[0].y,p[i].x-p[0].x); 44 sortf(1,tot); 45 ll ans=(n-1)*(n-2)*(n-3)/6; 46 ll tail=2,t=0; 47 for (int i=1;i<=tot;i++){ 48 while (cross(p[0],p[i],p[tail])>=0) {tail=tail%tot+1;t++;if (tail==i) break;} 49 ans-=t*(t-1)/2;t--; 50 } 51 return ans; 52 } 53 54 int main(){ 55 scanf("%lld",&n); 56 for (int i=1;i<=n;i++) scanf("%lld%lld",&a[i].x,&a[i].y); 57 ll an=0;for (int i=1;i<=n;i++) an+=calc(i); 58 ll b=n*(n-3)*(n-2)*(n-1)/24-an,c=2*b+an,d=(n-2)*n*(n-1)/6; 59 printf("%lf",(double)(c*4+(d-c)*3)/d); 60 return 0; 61 }
[BZOJ1911][BZOJ1912][BZOJ1913]APIO2010解题报告
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原文地址：http://www.cnblogs.com/mjy0724/p/4489591.html
