題目:VJ : UVa658
代碼實現:
#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<string>
#include<functional>
using namespace std;
struct Node{
int cost;
int id;
bool operator>(const Node& n)const{
return this->cost > n.cost;
}
};
int GetNext(int id,string prev,string post)
{
int mask = 1;
for(auto iter = prev.rbegin(); iter!=prev.rend(); ++iter){
if(*iter == '-' && !(id & mask)) //被WA了無數次
return -1;
if(*iter == '+' && (id & mask)) //位操作的慘痛教訓
return -1;
mask <<= 1;
}
mask = 1;
for(auto iter = post.rbegin(); iter!=post.rend(); ++iter){
if(*iter == '-'){
id = id|mask;
}
if(*iter == '+'){
id = id&(~mask);
}
mask <<= 1;
}
return id;
}
int main()
{
int n,m; //n-bugs數 m-patches數
int pid = 0; //產品編號
while(scanf("%d%d",&n,&m)==2 && n){
++pid;
vector<int> tm(m);
vector<string> prev(m);
vector<string> post(m);
//將m組patch輸入
for(int i = 0; i < m; ++i){
cin >> tm[i] >> prev[i] >> post[i];
}
//dijkstra算法
vector<bool> visited(1<<n,0);
int aim = (1<<n)-1;
priority_queue<Node,vector<Node>,greater<Node> > pq;
pq.push(Node{0,0});
bool flag = false;
while(!pq.empty()){
Node tn = pq.top(); pq.pop();
if(visited[tn.id])
continue;
visited[tn.id] = true;
if(tn.id == aim){
printf("Product %d\n",pid);
printf("Fastest sequence takes %d seconds.\n\n",tn.cost);
flag = true;
break;
}
for(int i = 0; i < m; ++i){
int nextid = GetNext(tn.id,prev[i],post[i]);
if(nextid != -1 && !visited[nextid]){
pq.push(Node{tm[i]+tn.cost,nextid});
}
}
}
if(!flag){
printf("Product %d\n",pid);
printf("Bugs cannot be fixed.\n\n");
}
}
return 0;
}
總結:採用dijkstra算法求最短路徑。被這道題卡了一天半,用的測試數據從題目給出的,到uDebug,再到自己寫程序隨機生成,無論怎麼測都沒問題,但一直WA;打開劉老師的代碼看,發現思路和我寫的相差無幾,代碼各個部分都能對上,然後通過代碼一段一段互換測試,定位出BUG在GetNext()函數,在第一個for循環中位操作後判斷出問題:原代碼爲(id & mask)==1,事實上需(id & mask)>0,因寫代碼時一直想着該位爲1,而該位若不在最低位,結果需>0纔可說明該位爲1。