2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛 H. Skiing（记忆化dfs）_综合

In this winter holiday, Bob has a plan for skiing at the mountain resort.

This ski resort has $M$ different ski paths and $N$ different flags situated at those turning points.

The $i$ -th path from the $S_i$ -th flag to the $T_i$ -th flag has length $L_i$ .

Each path must follow the principal of reduction of heights and the start point must be higher than the end point strictly.

An available ski trail would start from a flag, passing through several flags along the paths, and end at another flag.

Now, you should help Bob find the longest available ski trail in the ski resort.

Input Format

The first line contains an integer $T$ , indicating that there are $T$ cases.

In each test case, the first line contains two integers $N$ and $M$ where $\leq 10000$ and $\leq 100000$ as described above.

Each of the following $M$ lines contains three integers $S_i$ , $T_i$ , and $L_i~(0 < L_i < 1000)$ describing a path in the ski resort.

Output Format

For each test case, ouput one integer representing the length of the longest ski trail.

样例输入

样例输出

题解：

这一题是我a出来的第二个题目，一开始用bfs搜爆了内存，改进了TLE，然后就用dfs尝试了下居然过了。。。不知道别人怎么写的

代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
using namespace std;
#define INF 100861111
#define eps 1e-7
#define lson k*2
#define rson k*2+1
#define ll long long
struct edge
{int to;//该边通往的点int len;
}a[100005];
int p[10005][105];//存第i个点连接的第j条边是哪个
int num[10005];//存第i个点连接的边的数目
int r[10005];//存入度数目
int vis[10005];//存遍历情况
int dfs(int tar)//tar为当前点下标
{if(vis[tar])//如果已经遍历过了，直接返回已经保存的最大值{return vis[tar];}int i;int maxx=0;for(i=0;i<num[tar];i++){maxx=max(maxx,dfs(a[p[tar][i]].to)+a[p[tar][i]].len);//更新该点之后能遍历的最长长度}vis[tar]=maxx;//保存return maxx;//返回该最大值
}
int main()
{int i,j,k,n,m,test,x,y,d;scanf("%d",&test);while(test--){scanf("%d%d",&n,&m);for(i=1;i<=n;i++){num[i]=0;r[i]=0;vis[i]=0;}for(i=0;i<m;i++){scanf("%d%d%d",&x,&y,&d);a[i].len=d;a[i].to=y;p[x][num[x]]=i;num[x]++;r[y]++;}int maxx=0;for(i=1;i<=n;i++){if(r[i]==0)//从入度为0的点开始找{maxx=max(maxx,dfs(i));//更新最大值}}printf("%d\n",maxx);}return 0;
}