标签：des blog io os ar strong for 数据 div
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 10042 | Accepted: 4221 | Special Judge | ||
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input
3 5 4
Sample Output
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
应用时:10min
实际用时:57min
原因:六种操作都是不可逆转的.
思路:时间非常充裕到不需要建反边,数据太小
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=101;
int n,m;
typedef unsigned long long ull;
int A,B,C;
int vis[maxn][maxn];
int ans[maxn][maxn][maxn*maxn];
struct node{
int a,b;
node (int ta,int tb):a(ta),b(tb){}
};
void printop(int op){
switch(op){
case 0:
puts("FILL(1)");
break;
case 1:
puts("FILL(2)");
break;
case 2:
puts("POUR(1,2)");
break;
case 3:
puts("POUR(2,1)");
break;
case 4:
puts("DROP(1)");
break;
case 5:
puts("DROP(2)");
break;
}
}
void op(int &a,int &b,int op){
switch(op){
case 0:
a=A;
break;
case 1:
b=B;
break;
case 2:
if(b+a<=B){
b+=a;
a=0;
}
else {
a-=B-b;
b=B;
}
break;
case 3:
if(b+a<=A){
a+=b;
b=0;
}
else {
b-=A-a;
a=A;
}
break;
case 4:
a=0;
break;
case 5:
b=0;
break;
}
}
void bfs(){
queue <node> que;
que.push(node(0,0));
vis[0][0]=0;
while(!que.empty()){
node tp=que.front();que.pop();
int ta=tp.a;
int tb=tp.b;
if(tp.a==C||tp.b==C){
printf("%d\n",vis[tp.a][tp.b]);
for(int i=0;i<vis[ta][tb];i++){
int op=ans[ta][tb][i];
printop(op);
}
return ;
}
for(int i=0;i<6;i++){
int ta=tp.a;
int tb=tp.b;
op(ta,tb,i);
if(vis[ta][tb]==-1){
vis[ta][tb]=vis[tp.a][tp.b]+1;
for(int j=0;j<vis[tp.a][tp.b];j++){
ans[ta][tb][j]=ans[tp.a][tp.b][j];
}
ans[ta][tb][vis[tp.a][tp.b]]=i;
que.push(node(ta,tb));
}
}
}
puts("impossible");
}
int main(){
scanf("%d%d%d",&A,&B,&C);
memset(vis,-1,sizeof(vis));
bfs();
return 0;
}
标签：des blog io os ar strong for 数据 div
原文地址：http://www.cnblogs.com/xuesu/p/3992538.html
