[USACO3.3]亚瑟王的宫殿 Camelot 有点bug，求大神。调试3天了┭┮﹏┭┮
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> Run 13: Execution error: Your program did not produce an answer
that was judged as correct. The program stopped at 0.028 seconds;
it used 7132 KB of memory. At character number 2, your answer says
'0' while the correct answer says '1'.
Here are the respective outputs:
----- our output ---------
11
---- your output ---------
10
--------------------------
------ Data for Run 13 [length=24 bytes] ------
8 8
D 5
A 3 A 8 H 1 H 8
----------------------------
程序代码：/*
ID:luojiny1
LANG:C++
TASK:camelot
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
using namespace std;
const int maxR=31,maxC=28;
int R,C,n=0;
int kx,ky,nx[maxR*maxC],ny[maxR*maxC],d[maxC][maxR][maxC][maxR],ans=INF;
struct point{
int x,y,step;
};
void bfs(int sx,int sy){
int dx[8]={-1,1,2,2,1,-1,-2,-2};
int dy[8]={2,2,1,-1,-2,-2,-1,1};
bool vis[maxC][maxR]={0};
queue<point>Q;
Q.push((point){sx,sy,0});
while(!Q.empty()){
point a=Q.front();Q.pop();
int x=a.x,y=a.y;
if(vis[x][y])continue;
vis[x][y]=1;
d[sx][sy][x][y]=a.step;
for(int i=0;i<8;i++){
int newx=x+dx[i],newy=y+dy[i];
if(newx>=0&&newx<C&&newy>=0&&newy<R)Q.push((point){newx,newy,a.step+1});
}
}
}
int main()
{
// freopen("camelot.in","r",stdin);
// freopen("camelot.out","w",stdout);
memset(d,0x3f3f3f3f,sizeof(d));
char ch;
scanf("%d %d\n",&R,&C);
scanf("%c %d\n",&ch,&ky);
kx=ch-'A',ky--;
while((ch=getchar())!=EOF)if(ch>='A'&&ch<='Z'){
scanf("%d",&ny[n]);
ny[n]--,nx[n++]=ch-'A';
bfs(nx[n-1],ny[n-1]);
}
if(n==0){
printf("0\n");
return 0;
}
for(int dx=-1;dx<=1;dx++)
for(int dy=-1;dy<=1;dy++){
int newn=dx+kx,newm=dy+ky;
if(newn>=0&&newn<C&&newm>=0&&newm<R)bfs(newn,newm);}
for(int newx=0;newx<C;newx++)
for(int newy=0;newy<R;newy++)
{
int t=0;
for(int i=0;i<n;i++)
t+=d[nx[i]][ny[i]][newx][newy];
for(int dx=-1;dx<=1;dx++)
for(int dy=-1;dy<=1;dy++){
int newn=dx+kx,newm=dy+ky;
if(newn>=0&&newn<C&&newm>=0&&newm<R)
for(int i=0;i<n;i++)
ans=min(ans,t-d[nx[i]][ny[i]][newx][newy]+d[nx[i]][ny[i]][newn][newm]+abs(kx-newn)+abs(ky-newm)+min(d[newn][newm][newx][newy],abs(newn-newx)+abs(newm-newy)));
}
}
printf("%d\n",ans);
return 0;
}[此贴子已经被作者于2017-8-9 12:54编辑过]
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