HNU 1447 最长上升路径

最长上升路径

Time Limit: 3 Sec  Memory Limit: 128 MB Submissions: 103  Solved: 15

 

Description

给定n个顶点m条边的有向图，每一条边上都有一个正整数权值。一条有向路径被称为上升路径当且仅当除了路径上第一条边外，每一条表上的权值都严格大于路径上前一条边的权值。路径长度定义为路径所包含的边数，求给定图中最长上升路径长度。

 

Input

第一行是测试数据组数k。

接下来是k组测试数据。每组数据开始为n(n<=5000)和m(m<=100000)。接下来m行每行三个整数，表示图中一条边，格式为a b c，表示有一条从a到b的边，边权值为c。a和b是0到n-1的整数。

 

Output

每个测试数据输出一行。输出图中最长上升路径长度。

 

Sample Input

2
3 4
0 1 1
1 2 1
0 1 2
1 2 2
4 4
0 1 1
1 3 1
0 2 2
2 3 2

Sample Output

2
1

HINT

 

Source

 

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AC:

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <stack>
#include <ctime>
#include <algorithm>
#include <queue>

using namespace std;
#define INF 0x7fffffff
#define maxm 1001
#define mod 1000000007
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); i++)
#define re return
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define sqrt(x) sqrt(abs(x))
#define y0 y3487465
#define y1 y8687969
#define fill(x,y) memset(x,y,sizeof(x))

typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef double D;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef vector<vi> vvi;

template<class T> T abs(T x) { re x > 0 ? x : -x; }
const int maxn = 100015;
int n, m, t, k, x, l, r, s, sk,y;
int dp[maxn];
struct edge{
    int u, v, dist;
}edges[maxn];
bool cmp(edge a, edge b){ return a.dist < b.dist; }
pair<int, int>p;
int main(){
    scanf("%d", &t);
    while (t--){
        scanf("%d%d", &n, &m);
        memset(dp, 0, sizeof dp);
        stack<pair<int, int> >S;
        int ans = 0,i,j;
        for (int i = 0; i < m; i++)scanf("%d%d%d", &edges[i].u, &edges[i].v, &edges[i].dist);
        sort(edges, edges + m, cmp);
        for (int i = 0; i < m; i=j){
            for (j = i; edges[i].dist == edges[j].dist;j++)
            if (dp[edges[j].u] + 1>dp[edges[j].v]){
                p.first = edges[j].v; p.second = dp[edges[j].u] + 1;
                S.push(p);
            }
            while (!S.empty()){
                p = S.top(); S.pop();
                dp[p.first] = max(dp[p.first], p.second);
            }
        }
        for (int i = 0; i < n; i++)ans = max(ans, dp[i]);
        printf("%d\n", ans);
    }
    return 0;
}

 

TLE:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 100005
int n, m, t;
struct edge{
    int v, dist, next;
}edges[maxn];
int head[5005],dp[maxn];
void init(){
    for (int i = 0; i < maxn; i++)head[i] = -1;
}
void add(int u, int v, int dist, int id){
    edges[id].v = v; edges[id].dist = dist; edges[id].next = head[u]; head[u] = id;
}
int dfs(int u){
    if (dp[u] == -1){
        dp[u] = 1;
        for (int i = head[edges[u].v]; i != -1; i = edges[i].next)
            if (edges[i].dist>edges[u].dist)dp[u] = max(dp[u], dfs(i) + 1);
    }
    return dp[u];
}
int main(){
    scanf("%d", &t);
    while (t--){
        scanf("%d%d", &n, &m);
        init();
        memset(dp, -1, sizeof dp);
        for (int i = 0; i < m; i++){
            int a, b, len;
            scanf("%d%d%d", &a, &b, &len);
            add(a, b, len, i);
        }
        int ans = 0;
        for (int i = 0; i < m; i++)if (dp[i] == -1)ans = max(ans, dfs(i));
        printf("%d\n", ans);
    }
    return 0;
}