Luogu 4779(dijkstra+線段樹優化)(dijkstra+堆優化)
阿新 • • 發佈:2019-01-27
題意:模板題,求有向非負權圖的單源最短路
題解:
明說了要卡SPFA,所以只能dijkstra+資料結構優化,不管用堆還是線段樹,只有能到O(nlogn)就OK。
實測線段樹略快。
注意:每次“出隊”時將當前點賦值為INF(如果硬要做刪除操作就只有上平衡樹了233),線段樹在判斷“佇列為空”的邊界時直接判斷全域性最小值是否等於INF即可。
線段樹優化dijkstra:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r #define root 1,1,n const int N=1e5+4,M=2e5+4,INF=0x3f3f3f3f; int n,m,S; int head[N],etot=0,dis[N]; int mn[N<<2]; struct EDGE { int v,nxt,w; }e[M]; inline void adde(int u,int v,int w) { e[etot].nxt=head[u],e[etot].v=v,e[etot].w=w,head[u]=etot++; } inline int read() { int x=0;char c=getchar(); while (c<'0'||c>'9') c=getchar(); while (c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-'0',c=getchar(); return x; }/* struct Node { int id,dis; Node (int _id=0,int _dis=0):id(_id),dis(_dis) {} };*/ inline void pushup(int rt) { mn[rt]=min(mn[rt<<1],mn[rt<<1|1]); } void modify(int rt,int l,int r,int pos,int val) { if (l==r) { mn[rt]=val; return ; } int mid=l+r>>1; if (pos<=mid) modify(lson,pos,val); else modify(rson,pos,val); pushup(rt); } int query(int rt,int l,int r) { if (l==r) return l; int mid=l+r>>1; if (mn[rt<<1]<mn[rt<<1|1]) return query(lson); else return query(rson); } inline void dijkstra(int S) { memset(dis,INF,sizeof(dis)); dis[S]=0; modify(root,S,0); while (mn[1]^INF) { int p=query(root); modify(root,p,INF); for (int i=head[p];~i;i=e[i].nxt) { int v=e[i].v; if (dis[v]>dis[p]+e[i].w) { dis[v]=dis[p]+e[i].w; modify(root,v,dis[v]); } } } } int main() { // freopen("P4779.txt","r",stdin); memset(head,-1,sizeof(head)); memset(mn,INF,sizeof(mn)); n=read(),m=read(),S=read(); for (register int i=0;i<m;++i) { int u=read(),v=read(),w=read(); adde(u,v,w); } dijkstra(S); for (register int i=1;i<=n;++i) printf("%d ",dis[i]); puts(""); return 0; }
堆優化dijkstra:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; const int N=1e5+4,M=2e5+4; int n,m,S; int head[N],etot=0,dis[N]; struct EDGE { int v,nxt,w; }e[M]; struct Node { int id,dis; friend bool operator <(const Node &a,const Node &b) { return a.dis>b.dis; } }; inline void adde(int u,int v,int w) { e[etot].nxt=head[u],e[etot].v=v,e[etot].w=w,head[u]=etot++; } inline int read() { int x=0;char c=getchar(); while (c<'0'||c>'9') c=getchar(); while (c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-'0',c=getchar(); return x; } inline void dijkstra(int S) { priority_queue<Node > q; memset(dis,0x3f,sizeof(dis)); q.push((Node){S,dis[S]=0}); while (!q.empty()) { Node cur=q.top(); q.pop(); int p=cur.id; if (cur.dis^dis[p]) continue; for (int i=head[p];~i;i=e[i].nxt) { int v=e[i].v; if (dis[v]>dis[p]+e[i].w) { dis[v]=dis[p]+e[i].w; q.push((Node){v,dis[v]}); } } } } int main() { freopen("P4779.txt","r",stdin); memset(head,-1,sizeof(head)); n=read(),m=read(),S=read(); for (register int i=0;i<m;++i) { int u=read(),v=read(),w=read(); adde(u,v,w); } dijkstra(S); for (register int i=1;i<=n;++i) printf("%d ",dis[i]); puts(""); return 0; }