ACM-ICPC Live Archive Regionals 2014 >> 重拾2014上海区域赛
Show All Problems
| Problems Soluble |
|---|
| J - World Cup |
| I - Defeat the Enemy |
| B - Rotation |
src="https://icpcarchive.ecs.baylor.edu/external/71/7147.pdf" scrolling="auto" height="600" width="100%">
It is merely a problem about induction and deduction that trouble us mostly…
The (m+1)th team has to be considered specially.
/**
* Dec 1, 2015 8:20:12 PM
* PrjName: LA7147
* @semprathlon
*/
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException{
InputReader in=new InputReader(System.in);
PrintWriter out=new PrintWriter(System.out);
int T=in.nextInt(),cas=0;
while(T-->0){
long n=in.nextLong();
long m=in.nextLong();
long a=in.nextLong();
long b=in.nextLong();
long c=in.nextLong();
if (along ans1=Math.max(a, b)*(n-m-1);
long ans2=Math.min(c, b)*(m-1);
ans1+=Math.max(b*m, m/2*a+m/2*c+((m&1L)>0?Math.max(b, c):0));
ans2+=Math.min(b*(n-m), (n-m)/2*a+(n-m)/2*c+((n-m&1L)>0?Math.min(b, a):0));
out.println("Case #"+(++cas)+": "+ans1+" "+ans2);
}
out.flush();
out.close();
} src="https://icpcarchive.ecs.baylor.edu/external/71/7146.pdf" scrolling="auto" height="600" width="100%">
Bipartite graph matching turns out to exceed the time limit.Use greedy strategy as estimated.
Reflecting back upon the on-site period,the sort comparator can be considered correct.But the improper linear data structure failed to perform a “delete” operation.
Now use multiset instead.
Note that the optimal result doesn’t necessarily have to do with that earlier or later one of the tribes battles.
/*
* File: main.cpp
* Author: semprathlon
*
* Created on December 2, 2015, 7:31 PM
*/
#include int a,d;
scanf("%d%d",&a,&d);
c[i]=std::make_pair(a,d);
}
for(int i=0;iint a,d;
scanf("%d%d",&a,&d);
e[i]=std::make_pair(a,d);
}
std::sort(c,c+n,cmp1);
std::sort(e,e+m,cmp2);
// for(int i=0;i
// for(int i=0;i
int ans=n;
st.clear();
for(int i=0,j=0;ifor (;j=e[i].second;j++)
st.insert(c[j].second);
if (st.empty()){
ans=-1;break;
}
multiset<int>::iterator it=st.upper_bound(e[i].first);
if (it==st.end()){
st.erase(st.begin());
ans--;
}
else
st.erase(it);
}
printf("Case #%d: %d\n",++cas,ans);
}
return 0;
} src="https://icpcarchive.ecs.baylor.edu/external/71/7139.pdf" scrolling="auto" height="600" width="100%">
Terrific mistake occurred when reading the description of the problem.The path only add some value to the grids adjacent to it.
Also,pay attention to the square sum.
/**
* Dec 4, 2015 9:03:20 PM
* PrjName: LA7139
* @semprathlon
*/
import java.io.*;
import java.util.*;
public class Main {
static int[][] g;
static void add(int r,int c1,int c2,int v){
g[r][c1]+=v;
g[r][c2]-=v;
}
static void debug(int n,int m,PrintWriter out){
for(int i=1;i<=n+1;i++){
for(int j=1;j<=m+1;j++)
out.print(g[i][j]+" ");
out.println();
}
}
public static void main(String[] args) throws IOException{
InputReader in=new InputReader(System.in);
PrintWriter out=new PrintWriter(System.out);
int T=in.nextInt(),cas=0;
while(T-->0){
int n=in.nextInt();
int m=in.nextInt();
int k=in.nextInt();
g=new int[n+2][m+2];
int x=1,y=1;
for(int i=1;i<=k;i++){
String s=in.next();
int l=in.nextInt();
switch (s.charAt(0)) {
case 'R':
add(x,y,y+l,1);y+=l;break;
case 'L':
add(x, y-l, y, -1);y-=l;break;
case 'U':
x-=l;break;
case 'D':
x+=l;break;
}
}
long ans=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++){
g[i][j]+=g[i][j-1];
g[i][j]+=g[i-1][j];
g[i][j]-=g[i-1][j-1];
ans+=g[i][j]*g[i][j];
}
// debug(n, m, out);
out.println("Case #"+(++cas)+": "+ans);
}
out.flush();
out.close();
}
}