Skiing（ 2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛 ）

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

This ski resort has MM
different ski paths and NNdifferent
flags situated at those turning points.

The iii-th
path from the SiS_i-th
flag to the TiT_i​​-th
flag has length LiL_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 TTcases.

In each test case, the first line contains two integers NN
and MM
where 0<N≤100000 < N \leq 100000<N≤10000
and 0<M≤1000000 < M \leq 1000000<M≤100000
as described above.

Each of the following MMM
lines contains three integers SiS_i,TiT_i,
and Li (0<Li<1000)L_i~(0 < L_i < 1000)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

1
5 4
1 3 3
2 3 4
3 4 1
3 5 2

样例输出

6

题目来源

2017
ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛




1.拓扑排序。    
BFS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define inf 0x3f3f3f
#define ms(x) memset(x,0,sizeof(x))
#define mf(x) memset(x,-1,sizeof(x))
using namespace std;
const int N = 20002;
struct node{
int y,z;
node() {}
node(int ty, int tz): y(ty), z(tz){}
};
vector<node>q
;
int in
;
int f
;
int main(){
int T;
int n,m;
scanf("%d",&T);
while(T--){
queue<int>que;
ms(in);
ms(f);
int ans = 0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) q[i].clear();
for(int i=0;i<m;i++){
int x, y, z;
scanf("%d%d%d",&x,&y,&z);
q[x].push_back(node(y,z));
in[y]++;
}
for(int i=1;i<=n;i++){
if(!in[i]) que.push(i);
}
while(!que.empty()){
int cur = que.front();
que.pop();
for(int i=0;i<q[cur].size();i++){
int v = q[cur][i].y;
int w = q[cur][i].z;
f[v] = max(f[v],f[cur]+w);
ans = max(ans, f[v]);
in[v]--;
if(!in[v]) que.push(v);
}
}
cout<<ans<<endl;
}
return 0;
}