codeforces#327 div2
这场状态不好有点可惜，题目都不难，而且很好。。
A题：水题。
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef long long ll; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.000000000001; const double Pi=acos(-1.0); int l,p,q; int main() { //freopen("in.txt","r",stdin); while(cin>>l>>p>>q){ printf("%.5f ",l*1.0*p/(p+q)); } return 0; }
B题：水题。居然卡了，状态真差。。。
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef long long ll; const int maxn=200100; const int INF=(1<<29); const double EPS=0.000000000001; const double Pi=acos(-1.0); int n,q; int val[26]; char s[maxn]; int len; char u,v; char op[maxn]; int main() { //freopen("in.txt","r",stdin); while(cin>>n>>q){ scanf("%s",s); len=strlen(s); REP(i,0,25) val[i]=i; gets(op); while(q--){ //cin>>u>>v; gets(op); u=op[0],v=op[2]; u-='a';v-='a'; REP(i,0,25){ if(val[i]==u) val[i]=v; else if(val[i]==v) val[i]=u; } } REP(i,0,len-1){ s[i]=val[s[i]-'a']+'a'; } printf("%s ",s); } return 0; }
C题：水题。这个有思路，但是想着点到分心了。。
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef long long ll; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.000000000001; const double Pi=acos(-1.0); int n,a[maxn]; int main() { //freopen("in.txt","r",stdin); while(cin>>n){ REP(i,1,n) scanf("%d",&a[i]); int ans=0; for(int i=2;i<n;i++){ if(a[i]^a[i-1]&&a[i]^a[i+1]){ int l=i-1,r=i+1; while(l-1>=1&&a[l]^a[l-1]) l--; while(r+1<=n&&a[r]^a[r+1]) r++; //cout<<l<<" "<<r<<endl; if((r-l+1)&1){ REP(j,l,r) a[j]=a[r]; ans=max((r-l+1)/2,ans); } else{ int m=(l+r)/2; REP(j,l,m) a[j]=a[l]; REP(j,m+1,r) a[j]=a[r]; ans=max((r-l+1)/2-1,ans); } i=r; } } cout<<ans<<endl; REP(i,1,n){ if(i!=n) printf("%d ",a[i]); else printf("%d ",a[i]); } } return 0; }
D题：二分时间t。因为位移和运动轨迹并没有关系，位移也是满足矢量合成的，直接分解也可以。注意二分三分都写成for循环上1000次，不要带EPS，这样可以防止TLE。
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef long long ll; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); double X1,Y1,X2,Y2; double vm,t1; double vx,vy,wx,wy; double dist(double x,double y,double x0,double y0) { double tx=x-x0,ty=y-y0; return sqrt(tx*tx+ty*ty); } bool check(double t) { if(t<t1+EPS){ double x=X2-vx*t,y=Y2-vy*t; double d=dist(x,y,X1,Y1); //printf("%.1f %.1f %.1f %.1f %.1f %.1f %.1f ",x,y,d,X1,Y1,X2,Y2); if(vm*t>d-EPS) return 1; return 0; } else{ double x0=X2-vx*t1,y0=Y2-vy*t1; double x=x0-wx*(t-t1),y=y0-wy*(t-t1); double d=dist(x,y,X1,Y1); if(vm*t>d-EPS) return 1; return 0; } } double bin(double l,double r) { double res=r; REP(i,1,1000){ double m=(l+r)/2; if(check(m)) res=r=m; else l=m; } return res; } int main() { //freopen("in.txt","r",stdin); while(cin>>X1>>Y1>>X2>>Y2>>vm>>t1>>vx>>vy>>wx>>wy){ double t=bin(0,INF); printf("%.7f ",t); } return 0; }
E题：bfs预处理出每个点到三个国家的距离即可，然后路的长度就是三个距离直接加。
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; typedef long long ll; const int maxn=1100; const int INF=(1<<29); const double EPS=0.0000001; const double Pi=acos(-1.0); int n,m; char ch[maxn][maxn]; int d[3][maxn][maxn]; bool vis[maxn][maxn]; int dx[]={-1,1,0,0}; int dy[]={0,0,-1,1}; struct Node { int x,y; int step; }; void bfs(char c) { memset(d[c-'1'],-1,sizeof(d[c-'1'])); MS0(vis); queue<Node> q; REP(i,1,n){ REP(j,1,m){ if(ch[i][j]==c) q.push({i,j,0}),vis[i][j]=1; } } while(!q.empty()){ Node u=q.front(); q.pop(); d[c-'1'][u.x][u.y]=u.step; REP(i,0,3){ int nx=u.x+dx[i],ny=u.y+dy[i]; if(ch[nx][ny]=='#'||vis[nx][ny]) continue; vis[nx][ny]=1; q.push({nx,ny,u.step+1}); } } } void debug(char c) { cout<<endl; REP(i,1,n){ REP(j,1,m){ printf("%2d",d[c][i][j]); } cout<<endl; } } void deal() { int r1=INF,r2=INF; REP(i,1,n){ REP(j,1,m){ if(ch[i][j]=='1'){ r1=min(d[1][i][j],r1); r2=min(d[2][i][j],r2); } } } REP(i,1,n) REP(j,1,m) if(ch[i][j]=='1') d[1][i][j]=r1,d[2][i][j]=r2; r1=INF,r2=INF; REP(i,1,n){ REP(j,1,m){ if(ch[i][j]=='2'){ r1=min(d[0][i][j],r1); r2=min(d[2][i][j],r2); } } } REP(i,1,n) REP(j,1,m) if(ch[i][j]=='2') d[0][i][j]=r1,d[2][i][j]=r2; r1=INF,r2=INF; REP(i,1,n){ REP(j,1,m){ if(ch[i][j]=='3'){ r1=min(d[0][i][j],r1); r2=min(d[1][i][j],r2); } } } REP(i,1,n) REP(j,1,m) if(ch[i][j]=='3') d[0][i][j]=r1,d[1][i][j]=r2; } int main() { // freopen("in.txt","r",stdin); while(cin>>n>>m){ memset(ch,'#',sizeof(ch)); REP(i,1,n){ REP(j,1,m) cin>>ch[i][j]; } bfs('1');bfs('2');bfs('3'); deal(); //debug(0);debug(1);debug(2); int ans=INF; REP(i,1,n){ REP(j,1,m){ bool tag=1; REP(k,0,2){ if(d[k][i][j]==-1) tag=0; } if(!tag) continue; int tmp=0; tmp=d[0][i][j]+d[1][i][j]+d[2][i][j]-3+1; ans=min(ans,tmp); } } if(ans==INF) puts("-1"); else cout<<ans<<endl; } return 0; }
做题稳定性有待提升。