T1 貨物運輸弱化版

題解：

倒著跑最短路就行
沒仔細看題涼涼
code:

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #define F(i,a,b) for(register int i=(a);i<=(b);i++) using namespace std;int rd() {int x=0,f=1; char c=getchar();while(!isdigit(c)) {if(c=='-')f=-f; c=getchar();}while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();return x*f; }const int N=5010; int n,cnt,tot; int nxt[N],head[N],to[N],qu[N<<3],dp[N],vis[N];void bfs(int x) {//SPFAint h=0,t=1;qu[h]=x;while(h<=t) {int u=qu[h++],ret=0;vis[u]=0;if(u<27) {ret=dp[u]/19;if(ret*19!=dp[u]) ret++;} else ret=1;for(int i=head[u];i;i=nxt[i]) {int v=to[i];if(dp[v]>dp[u]+ret) {dp[v]=dp[u]+ret;if(!vis[dp[v]]) qu[t++]=v,vis[v]++;}}} }#define add(a,b) nxt[++cnt]=head[a],to[cnt]=b,head[a]=cnt #define insert(a,b) add(a,b),add(b,a) int main() {freopen("toll.in","r",stdin);freopen("toll.out","w",stdout);while(scanf("%d",&n),n!=-1) {cnt=0;tot++;memset(vis,0,sizeof(vis));memset(dp,0x3f,sizeof(dp));memset(head,0,sizeof(head));char a,b;F(i,1,n) {scanf(" %c %c",&a,&b);insert(a-'A'+1,b-'A'+1);}getchar();int ret; scanf("%d ",&ret);scanf("%c %c",&a,&b);int T=a-'A'+1,S=b-'A'+1;dp[S]=ret; bfs(S);printf("Case %d: %d\n",tot,dp[T]);//考試時直接輸出dp[T]QAQ涼涼 }return 0; }

T2 冒泡排序

求沒有逆序對的最長序列——不就是LIS么？ ——高 Orz
code:

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> using namespace std; int rd() {int x=0,f=1; char c=getchar();while(!isdigit(c)) {if(c=='-')f=-f; c=getchar();}while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();return x*f; } const int N=100010; int n,top,da[N],dp[N]; int main() {freopen("sort.in","r",stdin);freopen("sort.out","w",stdout);n=rd();for(int i=1;i<=n;i++) da[i]=rd();for(int i=1;i<=n;i++) if(dp[top]<da[i]) dp[++top]=da[i];else dp[lower_bound(dp+1,dp+1+top,da[i])-dp]=da[i];printf("%d",top);return 0; }

T3 room

比較水的狀壓
dp[i][j] i表示狀態 j表示不同顏色鑰匙數量
code:

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #define F(i,a,b) for(register int i=(a);i<=(b);i++) using namespace std;int rd() {int x=0,f=1; char c=getchar();while(!isdigit(c)) {if(c=='-')f=-f; c=getchar();}while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();return x*f; }int n,re,gr,wi,ans,mx; int rek[15],grk[15],wik[15],red[15],grd[15]; int dp[20000][3];bool jud(int a,int b) {int sum=0;int ret=dp[a-(1<<b)][0]-red[b+1]; if(ret<0) sum+=ret;ret=dp[a-(1<<b)][1]-grd[b+1]; if(ret<0) sum+=ret;return sum+dp[a-(1<<b)][2]>=0; }int main() {freopen("room.in","r",stdin);freopen("room.out","w",stdout);n=rd();F(i,1,n) red[i]=rd();F(i,1,n) grd[i]=rd();F(i,1,n) rek[i]=rd();F(i,1,n) grk[i]=rd();F(i,1,n) wik[i]=rd();dp[0][0]=rd(),dp[0][1]=rd(),dp[0][2]=rd();ans=dp[0][0]+dp[0][1]+dp[0][2];for(int i=0;i<(1<<n);i++) {for(int j=0;(i>>j);j++) if(((i>>j)&1)&&jud(i,j)) {dp[i][2]=dp[i-(1<<j)][2]+wik[j+1];if(dp[i-(1<<j)][0]-red[j+1]<0) {dp[i][2]+=dp[i-(1<<j)][0]-red[j+1];dp[i][0]=rek[j+1];} else dp[i][0]=dp[i-(1<<j)][0]-red[j+1]+rek[j+1];if(dp[i-(1<<j)][1]-grd[j+1]<0) {dp[i][2]+=dp[i-(1<<j)][1]-grd[j+1];dp[i][1]=grk[j+1];} else dp[i][1]=dp[i-(1<<j)][1]-grd[j+1]+grk[j+1];ans=max(ans,dp[i][0]+dp[i][1]+dp[i][2]);}}printf("%d",ans);return 0; }

轉載于:https://www.cnblogs.com/Menteur-Hxy/p/9268967.html

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