最短路 2 [HDU - 6714 ](dijkstra演算法)
阿新 • • 發佈:2020-07-23
最短路 2 [HDU - 6714 ](dijkstra演算法)
題目連結:https://vjudge.net/problem/HDU-6714
思路:
仔細分析可以得知: \(w[i][j]\)為\(i->j\)的最短路徑中不包含端點的最大編號節點(如果有多個最短路徑,選擇最大編號節點較小的那個。)
那麼\(w[i][j]\)可以在dijkstra演算法過程中動態規劃求出。
程式碼:
#include <iostream> #include <cstdio> #include <algorithm> #include <bits/stdc++.h> #define ALL(x) (x).begin(), (x).end() #define sz(a) int(a.size()) #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl #define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c)) #define du2(a,b) scanf("%d %d",&(a),&(b)) #define du1(a) scanf("%d",&(a)); using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;} ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;} void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;} inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;} void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}} void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}} const int maxn = 1005; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ #define DEBUG_Switch 0 int n, m; struct node { int to; ll val; node() {} node(int tt, ll vv) { to = tt; val = vv; } bool operator < (const node & b) const { return val > b.val; } }; std::vector<node> e[maxn]; ll dis[maxn][maxn]; int w[maxn][maxn]; void addedge(int a, int b, ll v) { e[a].push_back(node(b, v)); e[b].push_back(node(a, v)); } bool vis[maxn]; void init(int id, int n) { for (int i = 1; i <= n; ++i) { dis[id][i] = 1e18; vis[i] = 0; } } priority_queue<node> heap; void dijkstra(int id) { int strat = id; init(id, n); dis[id][strat] = 0ll; heap.push(node(strat, 0ll)); node temp; while (!heap.empty()) { temp = heap.top(); heap.pop(); if (vis[temp.to]) { continue; } else { vis[temp.to] = 1; } for (auto & x : e[temp.to]) { if (dis[id][temp.to] + x.val < dis[id][x.to]) { if (temp.to != id && temp.to != x.to) w[id][x.to] = max(temp.to, w[id][temp.to]); dis[id][x.to] = dis[id][temp.to] + x.val; heap.push(node(x.to, dis[id][x.to])); } else if (dis[id][temp.to] + x.val == dis[id][x.to]) { if (temp.to != id && temp.to != x.to) w[id][x.to] = min(w[id][x.to], max(w[id][temp.to], temp.to)); } } } } const int mod = 998244353; int main() { #if DEBUG_Switch freopen("C:\\code\\input.txt", "r", stdin); #endif //freopen("C:\\code\\output.txt","w",stdout); int t; t = readint(); while (t--) { n = readint(); m = readint(); repd(i, 1, n) { repd(j, 1, n) w[i][i] = 0; } repd(i, 1, m) { int x, y, val; x = readint(); y = readint(); val = readint(); addedge(x, y, val); w[x][y] = w[y][x] = 0; } repd(i, 1, n) { dijkstra(i); } int ans = 0; repd(i, 1, n) repd(j, 1, n) ans = (ans + w[i][j]) % mod; printf("%d\n", ans); repd(i, 1, n) e[i].clear(); } return 0; }