HDU4848 Wow! Such Conquering! —— dfs + 剪枝
2017-04-25 15:30
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4848
题解: 一开始读错题目。以为每个点只能访问一遍。其实只要每个点都有被访问就可以了。
首先是用弗洛伊德算法求出每两点之间的最短路。然后再用dfs搜索。注意剪枝,否则会超时。
代码如下;
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define ms(a, b) memset((a), (b), sizeof(a))
#define eps 0.0000001
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int maxn = 2000000+10;
const int mod = 1e9+7;
int dis[35][35],deadline[35];
int vis[35];
int n,ans;
void dfs(int u, int cnt, int time, int sum)
{
if(cnt==n)
{
ans = min(ans,sum);
return;
}
//剪枝:
if(sum+time*(n-cnt)>=ans) return;
for(int i = 2; i<=n; i++)
if(!vis[i] && time+dis[u][i]>deadline[i]) return;
for(int i = 2; i<=n; i++)
{
if(!vis[i])
{
vis[i] = 1;
dfs(i,cnt+1,time+dis[u][i], sum+time+dis[u][i]);
vis[i] = 0;
}
}
}
void solve()
{
for(int i = 1; i<=n; i++)
for(int j = 1; j<=n; j++)
scanf("%d",&dis[i][j]);
for(int i = 2; i<=n; i++)
scanf("%d",&deadline[i]);
for(int k = 1; k<=n; k++)//弗洛伊德算法
for(int i = 1; i<=n; i++)
for(int j = 1; j<=n; j++)
dis[i][j] = min(dis[i][j], dis[i][k]+dis[k][j]);
ans = INF;
ms(vis,0);
dfs(1,1,0,0);
if(ans!=INF)
printf("%d\n",ans);
else
puts("-1");
}
int main()
{
while(scanf("%d",&n)!=EOF){
solve();
}
return 0;
}
题解: 一开始读错题目。以为每个点只能访问一遍。其实只要每个点都有被访问就可以了。
首先是用弗洛伊德算法求出每两点之间的最短路。然后再用dfs搜索。注意剪枝,否则会超时。
代码如下;
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define ms(a, b) memset((a), (b), sizeof(a))
#define eps 0.0000001
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int maxn = 2000000+10;
const int mod = 1e9+7;
int dis[35][35],deadline[35];
int vis[35];
int n,ans;
void dfs(int u, int cnt, int time, int sum)
{
if(cnt==n)
{
ans = min(ans,sum);
return;
}
//剪枝:
if(sum+time*(n-cnt)>=ans) return;
for(int i = 2; i<=n; i++)
if(!vis[i] && time+dis[u][i]>deadline[i]) return;
for(int i = 2; i<=n; i++)
{
if(!vis[i])
{
vis[i] = 1;
dfs(i,cnt+1,time+dis[u][i], sum+time+dis[u][i]);
vis[i] = 0;
}
}
}
void solve()
{
for(int i = 1; i<=n; i++)
for(int j = 1; j<=n; j++)
scanf("%d",&dis[i][j]);
for(int i = 2; i<=n; i++)
scanf("%d",&deadline[i]);
for(int k = 1; k<=n; k++)//弗洛伊德算法
for(int i = 1; i<=n; i++)
for(int j = 1; j<=n; j++)
dis[i][j] = min(dis[i][j], dis[i][k]+dis[k][j]);
ans = INF;
ms(vis,0);
dfs(1,1,0,0);
if(ans!=INF)
printf("%d\n",ans);
else
puts("-1");
}
int main()
{
while(scanf("%d",&n)!=EOF){
solve();
}
return 0;
}
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