模拟+链表+DP
Orz AK爷faebdc
A
Growin要跟全部的n个人握手共2n杯香槟，再加上每对关系的两杯香槟，直接统计邻接矩阵中1的个数，再加2n就是answer
1 //BestCoder 42 A
2 #include<vector>
3 #include<cstdio>
4 #include<cstring>
5 #include<cstdlib>
6 #include<iostream>
7 #include<algorithm>
8 #define rep(i,n) for(int i=0;i<n;++i)
9 #define F(i,j,n) for(int i=j;i<=n;++i)
10 #define D(i,j,n) for(int i=j;i>=n;--i)
11 #define pb push_back
12 using namespace std;
13 inline int getint(){
14 int v=0,sign=1; char ch=getchar();
15 while(ch<‘0‘||ch>‘9‘){ if (ch==‘-‘) sign=-1; ch=getchar();}
16 while(ch>=‘0‘&&ch<=‘9‘){ v=v*10+ch-‘0‘; ch=getchar();}
17 return v*sign;
18 }
19 const int N=1e5+10,INF=~0u>>2;
20 typedef long long LL;
21 /******************tamplate*********************/
22 int n,a[100][100];
23 int main(){
24 #ifndef ONLINE_JUDGE
25 freopen("A.in","r",stdin);
26 freopen("A.out","w",stdout);
27 #endif
28 while(scanf("%d",&n)!=EOF){
29 memset(a,0,sizeof a);
30 LL cnt=0;
31 F(i,1,n) F(j,1,n){
32 a[i][j]=getint();
33 if (a[i][j]) cnt++;
34 }
35 printf("%lld\n",cnt+2*n);
36 }
37 return 0;
38 }
B
我一开始写了个以height为第一关键字，下标为第二关键字的set……每次直接lower_bound……然而TLE了
后来改了改离散化+链表模拟……结果FST了
查了半天想不出算法哪里错了……然后把10W的数组改成100W……过了
以后这种空间复杂度为线性的题目……在允许的情况下还是尽量开大点好了……
1 //BestCoder 42 B
2 #include<vector>
3 #include<cstdio>
4 #include<cstdlib>
5 #include<cstring>
6 #include<iostream>
7 #include<algorithm>
8 #define rep(i,n) for(int i=0;i<n;++i)
9 #define F(i,j,n) for(int i=j;i<=n;++i)
10 #define D(i,j,n) for(int i=j;i>=n;--i)
11 using namespace std;
12
13 int getint(){
14 int v=0,sign=1; char ch=getchar();
15 while(ch<‘0‘||ch>‘9‘) {if (ch==‘-‘) sign=-1; ch=getchar();}
16 while(ch>=‘0‘&&ch<=‘9‘) {v=v*10+ch-‘0‘; ch=getchar();}
17 return v*sign;
18 }
19 typedef long long LL;
20 const int N=1000100,INF=~0u>>2;
21 /*******************template********************/
22 int n,m,h[N],q[N],c[N<<1],head[N],v[N],nxt[N],cnt;
23 void add(int x,int y){
24 v[++cnt]=y; nxt[cnt]=head[x]; head[x]=cnt;
25 }
26 int main(){
27 #ifndef ONLINE_JUDGE
28 freopen("B.in","r",stdin);
29 // freopen("output.txt","w",stdout);
30 #endif
31 while(scanf("%d%d",&n,&m)!=EOF){
32 F(i,1,n) c[i]=h[i]=getint();
33 F(i,1,m) c[n+i]=q[i]=getint();
34 memset(head,0,sizeof head); cnt=0;
35 sort(c+1,c+n+m+1);
36 int num=unique(c+1,c+n+m+1)-c-1;
37 D(i,n,1){
38 h[i]=lower_bound(c+1,c+num+1,h[i])-c;
39 add(h[i],i);
40 }
41 F(i,1,m){
42 q[i]=lower_bound(c+1,c+num+1,q[i])-c;
43 if (head[q[i]]==0) puts("-1");
44 else{
45 printf("%d\n",v[head[q[i]]]);
46 head[q[i]]=nxt[head[q[i]]];
47 }
48 }
49 }
50 return 0;
51 }
C
直接背包DP吧……本来还想：会不会有人直接把ans的初值设为w[1][1]呢？然而w[1][1]>k？是不是可以hack一下……然而看了样例我发现我想多了……这场比赛的hack果然很少……
1 //BestCoder 42 C
2 #include<queue>
3 #include<set>
4 #include<vector>
5 #include<cstdio>
6 #include<cstring>
7 #include<cstdlib>
8 #include<iostream>
9 #include<algorithm>
10 #define rep(i,n) for(int i=0;i<n;++i)
11 #define F(i,j,n) for(int i=j;i<=n;++i)
12 #define D(i,j,n) for(int i=j;i>=n;--i)
13 #define pb push_back
14 using namespace std;
15 inline int getint(){
16 int v=0,sign=1; char ch=getchar();
17 while(ch<‘0‘||ch>‘9‘){ if (ch==‘-‘) sign=-1; ch=getchar();}
18 while(ch>=‘0‘&&ch<=‘9‘){ v=v*10+ch-‘0‘; ch=getchar();}
19 return v*sign;
20 }
21 const int N=110,INF=~0u>>2;
22 typedef long long LL;
23 /******************tamplate*********************/
24 int n,m,K,w[N][N];
25 bool f[N][N][N];
26 struct node{
27 int x,y,v;
28 };
29 queue<node>Q;
30 int main(){
31 #ifndef ONLINE_JUDGE
32 freopen("C.in","r",stdin);
33 freopen("C.out","w",stdout);
34 #endif
35 while(scanf("%d%d%d",&n,&m,&K)!=EOF){
36 memset(w,0,sizeof w);
37 memset(f,0,sizeof f);
38 F(i,1,n) F(j,1,m) w[i][j]=getint();
39 Q.push((node){1,1,0});
40 int ans=0;
41 if (w[1][1]<=K){
42 ans=w[1][1];
43 Q.push((node){1,1,w[1][1]});
44 }
45 while(!Q.empty()){
46 int x=Q.front().x,y=Q.front().y,v=Q.front().v; Q.pop();
47 if (x<n){
48 if (!f[x+1][y][v]){
49 f[x+1][y][v]=1;
50 Q.push((node){x+1,y,v});
51 }
52 if (v+w[x+1][y]<=K && !f[x+1][y][v+w[x+1][y]]){
53 f[x+1][y][v+w[x+1][y]]=1;
54 Q.push((node){x+1,y,v+w[x+1][y]});
55 ans=max(ans,v+w[x+1][y]);
56 }
57 }
58 if (y<m){
59 if (!f[x][y+1][v]){
60 f[x][y+1][v]=1;
61 Q.push((node){x,y+1,v});
62 }
63 if (v+w[x][y+1]<=K && !f[x][y+1][v+w[x][y+1]]){
64 f[x][y+1][v+w[x][y+1]]=1;
65 Q.push((node){x,y+1,v+w[x][y+1]});
66 ans=max(ans,v+w[x][y+1]);
67 }
68 }
69 }
70 printf("%d\n",ans);
71 }
72 return 0;
73 }
时间： 2024-08-25 05:25:37