抽空在vjudge上做了这套题。剩下FZU 2208数论题不会。
FZU 2205
这是个想法题，每次可以在上一次基础上加上边数/2的新边。
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 #include <string>
5 #include <string.h>
6 #include <stdio.h>
7 #include <queue>
8 #include <stack>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <ctime>
13 #include <cassert>
14 #include <sstream>
15 using namespace std;
16
17 const int N=2001;
18
19 int f[N];
20 int main () {
21 f[1]=0;
22 for (int i=2;i<=1000;i++) {
23 f[i]=f[i-1]+i/2;
24 }
25 int T;
26 cin>>T;
27 while (T--) {
28 int n;
29 cin>>n;
30 cout<<f[n]<<endl;
31 }
32 return 0;
33 }
FZU 2206
乍看不知道什么东西，直接在机器上跑几个数字可以试出来结论。
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 #include <string>
5 #include <string.h>
6 #include <stdio.h>
7 #include <queue>
8 #include <stack>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <ctime>
13 #include <cassert>
14 #include <sstream>
15 using namespace std;
16
17 const int N=501;
18 const int INF=0x3f3f3f3f;
19
20
21 int main () {
22 //freopen("out.txt","r",stdin);
23 int T;
24 scanf("%d",&T);
25 for (int cas=1;cas<=T;cas++) {
26 long long n;
27 scanf("%I64d",&n);
28 if (n<20150001LL) cout<<n+2014<<endl;
29 else cout<<20152014<<endl;
30 }
31 return 0;
32 }
FZU 2207
会用倍增算LCA的话，这题也肯定会做。
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 #include <string>
5 #include <string.h>
6 #include <stdio.h>
7 #include <queue>
8 #include <stack>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <ctime>
13 #include <cassert>
14 #include <sstream>
15 using namespace std;
16
17 const int N=2001;
18
19 vector<int>e[N];
20 int fa[N][13];
21 int dep[N];
22 void dfs(int u,int f,int d) {
23 dep[u]=d;
24 for (int i=0;i<e[u].size();i++) {
25 int v=e[u][i];
26 if (v==f) continue;
27 dfs(v,u,d+1);
28 fa[v][0]=u;
29 }
30 }
31
32 void calc(int n) {
33 for (int j=1;j<=12;j++) {
34 for (int i=1;i<=n;i++) {
35 fa[i][j]=fa[fa[i][j-1]][j-1];
36 }
37 }
38 }
39
40 int kthA(int u,int k) {
41 for (int i=12;i>=0;i--) {
42 if (k>=(1<<i)){
43 k-=(1<<i);
44 u=fa[u][i];
45 }
46 }
47 return u;
48 }
49
50 int lca(int u,int v) {
51 if (dep[u]<dep[v]) swap(u,v);
52 int d=dep[u]-dep[v];
53 u=kthA(u,d);
54 if (u==v) return u;
55 for (int i=12;i>=0;i--) {
56 if (fa[u][i]==fa[v][i])
57 continue;
58 u=fa[u][i];
59 v=fa[v][i];
60 }
61 return fa[u][0];
62 }
63 int main () {
64 int T;
65 scanf("%d",&T);
66 while (T--) {
67 int n,m;
68 scanf("%d %d",&n,&m);
69 for (int i=0;i<=n;i++)
70 e[i].clear();
71 for (int i=1;i<n;i++) {
72 int u,v;
73 scanf("%d %d",&u,&v);
74 e[u].push_back(v);
75 e[v].push_back(u);
76 }
77 dfs(1,-1,1);
78 calc(n);
79 static int cas=1;
80 printf("Case #%d:\n",cas++);
81 while (m--) {
82 int u,v,k;
83 scanf("%d %d %d",&u,&v,&k);
84 int ans=lca(u,v);
85 if (k==1) {
86 printf("%d\n",u);
87 continue;
88 }
89 else if (k==dep[u]-dep[ans]+dep[v]-dep[ans]+1) {
90 printf("%d\n",v);
91 continue;
92 }
93 if (dep[u]-dep[ans]+1>=k) {
94 int ret=kthA(u,k-1);
95 printf("%d\n",ret);
96 }
97 else {
98 int to=dep[u]-dep[ans]+dep[v]-dep[ans]+1-k;
99 int ret=kthA(v,to);
100 printf("%d\n",ret);
101 }
102 }
103 }
104 return 0;
105 }
FZU 2208
数论题，不会
FZU 2209
分层图。
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 #include <string>
5 #include <string.h>
6 #include <stdio.h>
7 #include <queue>
8 #include <stack>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <ctime>
13 #include <cassert>
14 #include <sstream>
15 using namespace std;
16
17 const int N=501;
18 const int INF=0x3f3f3f3f;
19 int get(int n,int u,int k) {
20 return k*n+u;
21 }
22 struct Edge {
23 int to,next,len;
24 Edge() {}
25 Edge(int _to,int _next,int _len):to(_to),next(_next),len(_len) {}
26 } edge[250000];
27 int idx=1,head[N];
28 inline void addedge(int u,int v,int l) {
29 edge[++idx]=Edge(v,head[u],l);
30 head[u]=idx;
31 }
32 int dis[N][N],in[N];
33 bool vis[N];
34 bool spfa(int s,int n,int *dis) {
35 fill(dis,dis+N,INF);
36 memset(vis,false,sizeof(vis));
37 memset(in,0,sizeof(in));
38 dis[s]=0;
39 vis[s]=true;
40 queue<int> que;
41 que.push(s);
42 in[s]=1;
43 while (!que.empty()) {
44 int u=que.front();
45 que.pop();
46 vis[u]=false;
47 for (int k=head[u];k;k=edge[k].next) {
48 int v=edge[k].to;
49 if (dis[v]>dis[u]+edge[k].len) {
50 dis[v]=dis[u]+edge[k].len;
51 if (!vis[v]) {
52 vis[v]=true;
53 in[v]++;
54 if (in[v]>n) return false;
55 que.push(v);
56 }
57 }
58 }
59 }
60 return true;
61 }
62 int main () {
63 //freopen("out.txt","r",stdin);
64 int T;
65 scanf("%d",&T);
66 for (int cas=1;cas<=T;cas++) {
67 int n,m,k;
68 scanf("%d %d %d",&n,&m,&k);
69 idx=1;memset(head,0,sizeof head);
70 for (int i=1;i<=m;i++) {
71 int u,v;
72 scanf("%d %d",&u,&v);
73 int d;
74 for (int i=0;i<24;i++) {
75 scanf("%d",&d);
76 int uu=get(n,u,i);
77 int vv=get(n,v,(i+d)%24);
78 addedge(uu,vv,d);
79 uu=get(n,v,i);
80 vv=get(n,u,(i+d)%24);
81 addedge(uu,vv,d);
82 }
83 }
84 for (int i=0;i<24;i++) {
85 spfa(get(n,1,i),n*24,dis[i]);
86 }
87 printf("Case #%d:",cas);
88 while (k--) {
89 int v,s;
90 scanf("%d %d",&v,&s);
91 //cout<<"from "<<get(n,1,s)<<endl;
92 int ret=0x3f3f3f3f;
93 for (int i=0;i<24;i++) {
94 int p=get(n,v,i);
95 ret=min(ret,dis[s][p]);
96 }
97 if (ret==0x3f3f3f3f) ret=-1;
98 printf(" %d",ret);
99 }
100 puts("");
101 }
102 return 0;
103 }
FZU 2210
建立一个虚拟节点，连向所有粮仓，枚举所有禁止的城市，从虚拟节点dfs，更新答案。
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 #include <string>
5 #include <string.h>
6 #include <stdio.h>
7 #include <queue>
8 #include <stack>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <ctime>
13 #include <cassert>
14 #include <sstream>
15 using namespace std;
16
17 const int N=2001;
18
19 vector<int>e[N];
20 int in[N];
21 bool vis[N];
22 int fob;
23 int cnt;
24 void dfs(int u) {
25 if (u==fob||vis[u]) return;
26 vis[u]=true;cnt++;
27 for (int i=0;i<e[u].size();i++) {
28 int v=e[u][i];
29 dfs(v);
30 }
31 }
32 int main () {
33 int n,m;
34 while (scanf("%d %d",&n,&m)!=EOF) {
35 for (int i=0;i<=n;i++){
36 e[i].clear();
37 in[i]=0;
38 }
39 for (int i=1;i<=m;i++) {
40 int u,v;
41 scanf("%d %d",&u,&v);
42 e[u].push_back(v);
43 in[v]++;
44 }
45 for (int i=1;i<=n;i++) {
46 if (in[i]==0)
47 e[0].push_back(i);
48 }
49 int ret=n+1,val=0;
50 for (int i=n;i>=1;i--) {
51 fob=i;
52 memset(vis,false,sizeof vis);
53 cnt=0;
54 dfs(0);
55 cnt--;
56 if (n-cnt>=val) {
57 val=n-cnt;
58 ret=i;
59 }
60 }
61 cout<<ret<<endl;
62 }
63 return 0;
64 }
FZU 2211
费用流，把所有蘑菇拆成两个点，根据时间关系，互相连边。农田的限制在于源点出来的流的大小。
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 #include <string>
5 #include <string.h>
6 #include <stdio.h>
7 #include <queue>
8 #include <stack>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <ctime>
13 #include <cassert>
14 #include <sstream>
15 using namespace std;
16
17 const int N=5010;
18 const int INF=0x3f3f3f3f;
19
20 struct Edge{
21 int to,next,f;
22 long long c;
23 Edge(){}
24 Edge(int _to,int _nxt,int _f,long long _c):to(_to),next(_nxt),f(_f),c(_c){}
25 }edge[N<<2];
26
27 int head[N],idx;
28 bool vis[N];
29 long long dis[N];
30 int pree[N],prev[N];
31 void addedge(int u,int v,int flow,long long cost){
32 edge[++idx]=Edge(v,head[u],flow,cost);
33 head[u]=idx;
34 edge[++idx]=Edge(u,head[v],0,-cost);
35 head[v]=idx;
36 }
37 bool spfa(int s,int e){
38 memset(vis,0,sizeof(vis));
39 memset(pree,-1,sizeof(pree));
40 memset(prev,-1,sizeof(prev));
41 for (int i=0;i<N;i++) dis[i]=~0ULL>>3;
42 dis[s]=0;
43 vis[s]=true;
44 queue<int>que;
45 que.push(s);
46 while (!que.empty()){
47 int cur=que.front();
48 que.pop();
49 vis[cur]=false;
50 for (int k=head[cur];k;k=edge[k].next){
51 if (edge[k].f){
52 int n=edge[k].to;
53 if (dis[n]>dis[cur]+edge[k].c){
54 dis[n]=dis[cur]+edge[k].c;
55 prev[n]=cur;
56 pree[n]=k;
57 if (!vis[n]){
58 vis[n]=true;
59 que.push(n);
60 }
61 }
62 }
63 }
64 }
65 if (dis[e]>=INF) return 0;
66 return 1;
67 }
68 long long minCostMaxFlow(int src,int sink){
69 long long cur,min_val;
70 long long ansf=0,ansc=0;
71 while (spfa(src,sink)){
72 cur=sink;
73 min_val=INF;
74 while (prev[cur]!=-1){
75 if (min_val>edge[pree[cur]].f)
76 min_val=edge[pree[cur]].f;
77 cur=prev[cur];
78 }
79 cur=sink;
80 while (prev[cur]!=-1){
81 edge[pree[cur]].f-=min_val;
82 edge[pree[cur]^1].f+=min_val;
83 cur=prev[cur];
84 }
85 ansc+=dis[sink]*min_val;
86 ansf+=min_val;
87 }
88 return ansc;
89 }
90
91 int s[N],t[N],v[N];
92 int main () {
93 int T;
94 scanf("%d",&T);
95 for (int cas=1;cas<=T;cas++) {
96 int n,m;
97 scanf("%d %d",&n,&m);
98 for (int i=1;i<=m;i++) {
99 scanf("%d %d %d",s+i,t+i,v+i);
100 }
101 idx=1;memset(head,0,sizeof head);
102 for (int i=1;i<=m;i++) {
103 for (int j=1;j<=m;j++) {
104 if (t[i]<s[j]) {
105 addedge(i+m,j,INF,0);
106 }
107 }
108 }
109 int s=m+m+1,t=m+m+1+1;
110 for (int i=1;i<=m;i++) {
111 addedge(0,i,INF,0);
112 addedge(i,i+m,1,-v[i]);
113 addedge(i+m,t,INF,0);
114 }
115 addedge(s,0,n,0);
116 long long ret=-minCostMaxFlow(s,t);
117 cout<<ret<<endl;
118 }
119 return 0;
120 }
时间： 2024-10-11 02:54:28