标签：
A
水
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <list> 5 #include <map> 6 #include <stack> 7 #include <vector> 8 #include <cstring> 9 #include <sstream> 10 #include <string> 11 #include <cmath> 12 #include <queue> 13 using namespace std; 14 #define clc(a,b) memset(a,b,sizeof(a)) 15 #define inf 0x3f3f3f3f 16 const int N=10010; 17 const int MOD = 1e9+7; 18 #define LL long long 19 void fre() { 20 freopen("in.txt","r",stdin); 21 } 22 inline int r() { 23 int x=0,f=1;char ch=getchar(); 24 while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1;ch=getchar();} 25 while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘;ch=getchar();}return x*f; 26 } 27 28 int main(){ 29 int T; 30 T=r(); 31 while(T--){ 32 int x,y; 33 int ans; 34 x=r(); 35 y=r(); 36 if(x%y!=0) 37 ans=x/y+1; 38 else 39 ans=x/y; 40 cout<<ans<<endl; 41 } 42 return 0; 43 }
C
最短路
反向建边，记录当前点序号最小的前驱
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <list> 5 #include <map> 6 #include <stack> 7 #include <vector> 8 #include <cstring> 9 #include <sstream> 10 #include <string> 11 #include <cmath> 12 #include <queue> 13 #include <bits/stdc++.h> 14 using namespace std; 15 #define clc(a,b) memset(a,b,sizeof(a)) 16 #define inf 0x3f3f3f3f 17 const int N=100010; 18 const int MOD = 1e9+7; 19 #define LL long long 20 void fre() { 21 freopen("in.txt","r",stdin); 22 } 23 inline int r() { 24 int x=0,f=1;char ch=getchar(); 25 while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1;ch=getchar();} 26 while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘;ch=getchar();}return x*f; 27 } 28 int n,m; 29 struct node{ 30 int v,c; 31 node(int vv,int cc){ 32 v=vv; 33 c=cc; 34 } 35 node(){} 36 bool operator <(const node &r) const{ 37 return c>r.c; 38 } 39 }; 40 41 struct edge{ 42 int v,cost; 43 edge(int vv=0,int ccost =0):v(vv),cost(ccost){} 44 }; 45 46 vector<edge>e[N]; 47 bool vis[N]; 48 int dist[N]; 49 int p[N]; 50 void dij(int start){ 51 clc(vis,false); 52 for(int i=0;i<=n+1;i++) dist[i]=inf; 53 priority_queue<node>q; 54 while(!q.empty()) q.pop(); 55 dist[start]=0; 56 q.push(node(start,0)); 57 node tmp; 58 while(!q.empty()){ 59 tmp=q.top(); 60 q.pop(); 61 int u=tmp.v; 62 if(vis[u]) continue; 63 vis[u]=true; 64 for(int i=0;i<e[u].size();i++){ 65 int v=e[u][i].v; 66 int cost=e[u][i].cost; 67 if(!vis[v]&&dist[v]>dist[u]+cost){ 68 dist[v]=dist[u]+cost; 69 p[v]=u; 70 q.push(node(v,dist[v])); 71 } 72 else if(!vis[v]&&dist[v]==dist[u]+cost){ 73 p[v]=min(p[v],u); 74 } 75 } 76 } 77 } 78 void add(int u,int v,int w){ 79 e[u].push_back(edge(v,w)); 80 } 81 82 void init(){ 83 for(int i=0;i<=n+1;i++){ 84 e[i].clear(); 85 } 86 clc(p,-1); 87 } 88 89 int main(){ 90 // fre(); 91 int T; 92 T=r(); 93 while(T--){ 94 n=r(),m=r(); 95 init(); 96 int u,v,w; 97 while(m--){ 98 u=r(),v=r(),w=r(); 99 add(v,u,w); 100 } 101 dij(n+1); 102 if(dist[0]>=inf){ 103 printf("-1\n"); 104 continue; 105 } 106 else if(p[0]==n+1){ 107 printf("0\n"); 108 continue; 109 } 110 else 111 printf("%d\n",p[0]); 112 } 113 return 0; 114 }
D
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <list> 5 #include <map> 6 #include <stack> 7 #include <vector> 8 #include <cstring> 9 #include <sstream> 10 #include <string> 11 #include <cmath> 12 #include <queue> 13 using namespace std; 14 #define clc(a,b) memset(a,b,sizeof(a)) 15 #define inf 0x3f3f3f3f 16 const int N=200010; 17 const int MOD = 1e9+7; 18 #define LL long long 19 void fre() { 20 freopen("in.txt","r",stdin); 21 } 22 inline int r() { 23 int x=0,f=1;char ch=getchar(); 24 while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1;ch=getchar();} 25 while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘;ch=getchar();}return x*f; 26 } 27 struct node{ 28 int s,w,num; 29 }p[N]; 30 int n,R,q; 31 node a[N],b[N]; 32 33 bool cmp(const node &a,const node &b){ 34 return (a.s==b.s)?(a.num<b.num):(a.s>b.s); 35 } 36 37 void work(){ 38 int ai=1,bi=1; 39 for(int i=1;i<=2*n;i+=2){ 40 if(p[i].w>p[i+1].w){ 41 p[i].s++; 42 a[ai++]=p[i]; 43 b[bi++]=p[i+1]; 44 } 45 else{ 46 p[i+1].s++; 47 a[ai++]=p[i+1]; 48 b[bi++]=p[i]; 49 } 50 } 51 int i=1,j=1,k=1; 52 while(i<ai&&j<bi){ 53 if(cmp(a[i],b[j])){ 54 p[k++]=a[i++]; 55 } 56 else 57 p[k++]=b[j++]; 58 } 59 while(i<ai) p[k++]=a[i++]; 60 while(j<bi) p[k++]=b[j++]; 61 } 62 63 int main(){ 64 // fre(); 65 int T; 66 T=r(); 67 while(T--){ 68 n=r(),R=r(),q=r(); 69 for(int i=1;i<=2*n;i++){ 70 int x; 71 x=r(); 72 p[i].s=x; 73 p[i].num=i; 74 } 75 for(int i=1;i<=2*n;i++){ 76 int x; 77 x=r(); 78 p[i].w=x; 79 } 80 sort(p+1,p+1+2*n,cmp); 81 for(int i=1;i<=R;i++){ 82 work(); 83 } 84 printf("%d\n",p[q].num); 85 } 86 return 0; 87 }
F
dp记忆话搜索
dp[i][j][k][inx][last]:当前三种水果分别有i j k个的时候且当前放第inx类的水果，持续了last天的方案数
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <list> 5 #include <map> 6 #include <stack> 7 #include <vector> 8 #include <cstring> 9 #include <sstream> 10 #include <string> 11 #include <cmath> 12 #include <queue> 13 using namespace std; 14 #define clc(a,b) memset(a,b,sizeof(a)) 15 #define inf 0x3f3f3f3f 16 const int N=10010; 17 const int MOD = 1e9+7; 18 #define LL long long 19 void fre() { 20 freopen("in.txt","r",stdin); 21 } 22 inline int r() { 23 int x=0,f=1;char ch=getchar(); 24 while(ch>‘9‘||ch<‘0‘) {if(ch==‘-‘) f=-1;ch=getchar();} 25 while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘;ch=getchar();}return x*f; 26 } 27 28 int num; 29 int dp[51][51][51][3][51]; 30 int a[3],b[3]; 31 32 int dfs(int n,int a0,int a1,int a2,int inx,int last){ 33 LL ans=0; 34 if(dp[a0][a1][a2][inx][last]!=-1) return dp[a0][a1][a2][inx][last]; 35 if(n==num) return dp[a0][a1][a2][inx][last]=1; 36 if(inx==-1){ 37 if(b[0]>0&&a0>=1) ans+=dfs(n+1,a0-1,a1,a2,0,1); 38 if(b[1]>0&&a1>=1) ans+=dfs(n+1,a0,a1-1,a2,1,1); 39 if(b[2]>0&&a2>=1) ans+=dfs(n+1,a0,a1,a2-1,2,1); 40 } 41 else{ 42 if(inx==0){ 43 if(last+1<=b[0]&&a0>=1) ans+=dfs(n+1,a0-1,a1,a2,0,last+1); 44 if(b[1]>0&&a1>=1) ans+=dfs(n+1,a0,a1-1,a2,1,1); 45 if(b[2]>0&&a2>=1) ans+=dfs(n+1,a0,a1,a2-1,2,1); 46 } 47 else if(inx==1){ 48 if(last+1<=b[1]&&a1>=1) ans+=dfs(n+1,a0,a1-1,a2,1,last+1); 49 if(b[0]>0&&a0>=1) ans+=dfs(n+1,a0-1,a1,a2,0,1); 50 if(b[2]>0&&a2>=1) ans+=dfs(n+1,a0,a1,a2-1,2,1); 51 } 52 else{ 53 if(last+1<=b[2]&&a2>=1) ans+=dfs(n+1,a0,a1,a2-1,2,last+1); 54 if(b[0]>0&&a0>=1) ans+=dfs(n+1,a0-1,a1,a2,0,1); 55 if(b[1]>0&&a1>=1) ans+=dfs(n+1,a0,a1-1,a2,1,1); 56 } 57 } 58 ans%=MOD; 59 return dp[a0][a1][a2][inx][last]=ans; 60 } 61 62 int main(){ 63 int T; 64 T=r(); 65 while(T--){ 66 clc(dp,-1); 67 for(int i=0;i<3;i++){ 68 // scanf("%d",&a[i]); 69 int x; 70 x=r(); 71 a[i]=x; 72 } 73 for(int i=0;i<3;i++){ 74 // scanf("%d",&b[i]); 75 int x; 76 x=r(); 77 b[i]=x; 78 } 79 num=a[0]+a[1]+a[2]; 80 printf("%d\n",dfs(0,a[0],a[1],a[2],-1,0)); 81 } 82 return 0; 83 }
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原文地址：http://www.cnblogs.com/ITUPC/p/5573854.html
