HDU 1874 畅通工程续 戳我直达
题意:求s到t的最短路径。最简单的最短路径题。
做法:Dijkstra 或 floyd
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#include <bits/stdc++.h> #define scf0(a) scanf("%s",&a) #define scf1(a) scanf("%d",&a) #define scf2(a,b) scanf("%d%d",&a,&b) #define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c) #define MEM(a,b) memset(a,b,sizeof(a)) #define pii pair<int,int> #define pdd pair<double,double> #define LL long long using namespace std; const int INF = 0x3f3f3f3f; const double eps = 1e-8; const int maxn = 200 + 10; int mp[maxn][maxn]; int n, m, s, t, u, v, w; int Dijkstra( int s, int t) { //初始化 int dist[maxn]; bool vis[maxn]; MEM(vis, false ); for ( int i = 0; i < n; i++) dist[i] = mp[s][i]; dist[s] = 0; vis[s] = true ; //循环n - 1次 for ( int j = 0; j < n - 1; j++) { //初始化 int minn = INF, pos; //找最小点 for ( int i = 0; i < n; i++) { if (vis[i]) continue ; if (dist[i] < minn) pos = i, minn = dist[i]; } vis[pos] = true ; //更新最短路径 for ( int i = 0; i < n; i++) { if (vis[i]) continue ; dist[i] = min(dist[i], dist[pos] + mp[pos][i]); } } return dist[t] < INF ? dist[t] : -1; } int main() { while (~scf2(n, m)) { MEM(mp, INF); for ( int i = 0; i < m; i++) { scf3(u, v, w); if (w < mp[u][v]) mp[u][v] = mp[v][u] = w; //输入的小坑...两点之间可能不止两条路线 } scf2(s, t); printf ( "%d
" , Dijkstra(s, t) ); } } |
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#include <bits/stdc++.h> #define scf0(a) scanf("%s",&a) #define scf1(a) scanf("%d",&a) #define scf2(a,b) scanf("%d%d",&a,&b) #define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c) #define MEM(a,b) memset(a,b,sizeof(a)) #define pii pair<int,int> #define pdd pair<double,double> #define LL long long using namespace std; const int INF = 0x3f3f3f3f; const double eps = 1e-8; const int maxn = 200 + 10; int mp[maxn][maxn]; int n, m, s, t, u, v, w; int floyd( int s, int t) //无fuck说,不想懂 { for ( int k=0;k<n;k++) for ( int i=0;i<n;i++) for ( int j=0;j<n;j++) mp[i][j]=mp[j][i]=min(mp[i][j],mp[i][k]+mp[k][j]); return mp[s][t] < INF ? mp[s][t] : -1; } int main() { while (~scf2(n, m)) { MEM(mp, INF); for ( int i = 0; i < m; i++) { scf3(u, v, w); if (w < mp[u][v]) mp[u][v] = mp[v][u] = w; //输入的小坑...两点之间可能不止两条路线 } scf2(s, t); if (s == t) puts ( "0" ); //相同始终点,floyd最终得到的0 else printf ( "%d
" , floyd(s, t) ); } } |