Problem Description

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?ii-th path from the?Si?-th flag to the?Ti?-th flag has length?Li?.

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

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?0<N≤10000?and?0<M≤100000as described above.

Each of the following?M?lines contains three integers?Si?,?Ti?, and?Li??(0<Li?<1000)?describing a path in the ski resort.

Output

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

Sample?Input

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

???????Sample????????Output

6

題意：t 組數(shù)據(jù)，每組給出一個(gè) n 個(gè)點(diǎn) m 條邊的帶權(quán)有向圖，問整個(gè)圖的最長(zhǎng)路

思路：DAG 圖求最長(zhǎng)路，利用拓?fù)渑判騺斫鉀Q即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long #define Pair pair<int,int> const int MOD = 1E9+7; const int N = 50000+5; const int dx[] = {1,-1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;struct Node{int to,dis;Node(){}Node(int to,int dis):to(to),dis(dis){} }; vector<Node> G[N]; int in[N]; int dis[N]; int n,m;void topSort() {stack<int > S;for(int i=1; i<=n; i++)if(!in[i])S.push(i);while(!S.empty()) {int x=S.top();S.pop();for(int j=0; j<G[x].size(); j++) {int y=G[x][j].to;dis[y]=max(dis[y],dis[x]+G[x][j].dis);in[y]--;if(!in[y])S.push(y);}} }int main() {int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);memset(in,0,sizeof(in));memset(dis,0,sizeof(dis));for(int i=0; i<=n; i++)G[i].clear();for(int i=1; i<=m; i++) {int x,y,dis;scanf("%d%d%d",&x,&y,&dis);Node temp;temp.to=y;temp.dis=dis;in[y]++;G[x].push_back(temp);}topSort();int res=-INF;for(int i=1;i<=n;i++)res=max(res,dis[i]);printf("%d\n",res);}return 0; }

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總結(jié)

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