17-10-15模拟赛

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define inf 0x7fffffff
 5 #define MN 2005
 6 using namespace std;
 7 inline int in(){
 8     int x=0;bool f=0; char c;
 9     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
10     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
11     return f?-x:x;
12 }
13 int f[MN],a[MN],h[MN],pos[MN];
14 int n,cnt,res;
15 inline void dp(bool d){
16     memset(f,0,sizeof(f));for (int i=1;i<=n;++i){
17         if (d) for (int j=1;j<=cnt;++j) f[j]=min(f[j],(j-1)?f[j-1]:inf);
18         else for (int j=cnt;j;--j) f[j]=min(f[j],(j<cnt)?f[j+1]:inf);
19         for (int j=1;j<=cnt;++j) f[j]+=abs(a[pos[i]]-a[j]);
20     }for (int i=1;i<=cnt;++i) res=min(res,f[i]);
21 }
22 int main()
23 {
24     n=in();res=inf;for (int i=1;i<=n;++i) a[i]=h[i]=in();
25     sort(a+1,a+n+1);cnt=unique(a+1,a+n+1)-a-1;
26     for (int i=1;i<=n;++i) pos[i]=lower_bound(a+1,a+n+1,h[i])-a;
27     dp(0);dp(1);printf("%d",res);return 0;
28 }

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define inf 0x3f3f3f3f
 5 #define MM 100005
 6 #define MN 105 
 7 using namespace std;
 8 inline int in(){
 9     int x=0;bool f=0; char c;
10     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
11     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
12     return f?-x:x;
13 }
14 int f[MM],p[MN],c[MN],res=inf,n,m;
15 int main()
16 {
17     n=in();m=in();memset(f,0x3f,sizeof(f));f[0]=0;
18     for (int i=1;i<=n;++i) p[i]=in(),c[i]=in();
19     for (int i=1;i<=n;++i)
20     for (int j=p[i];j<=(m<<1);++j){
21         if (j<m) f[j]=min(f[j],f[j-p[i]]+c[i]);
22         else res=min(res,f[j-p[i]]+c[i]);
23     }printf("%d",res);return 0;
24 }

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define MN 10005
 5 using namespace std;
 6 inline int in(){
 7     int x=0;bool f=0; char c;
 8     for (;(c=getchar())<‘0‘||c>‘9‘;f=c==‘-‘);
 9     for (x=c-‘0‘;(c=getchar())>=‘0‘&&c<=‘9‘;x=(x<<3)+(x<<1)+c-‘0‘);
10     return f?-x:x;
11 }
12 struct rel{
13     int a,b;
14 }p[MN];
15 inline bool cmp(rel x,rel y){return (x.a==y.a)?x.b<y.b:x.a<y.a;}
16 inline bool comp(rel x,rel y){return (x.a==y.a)&&(x.b==y.b);}
17 int f[MN],n,l,h,m,sum=0,cnt=0;
18 int main()
19 {
20     n=in();l=in();h=in();m=in();f[1]+=h;f[n+1]-=h;
21     for (int i=1;i<=m;++i){
22         p[i].a=in();p[i].b=in();
23         if (p[i].a>p[i].b) swap(p[i].a,p[i].b);
24     }sort(p+1,p+m+1,cmp);cnt=unique(p+1,p+m+1,comp)-p-1;
25     for (int i=1;i<=cnt;++i) --f[p[i].a+1],++f[p[i].b];
26     for (int i=1;i<=n;++i) sum+=f[i],printf("%d\n",sum);return 0;
27 }