最小生成树Prim算法模板

来源：互联网发布：詹姆斯深蹲力量数据编辑：程序博客网时间：2024/05/22 07:49

不带堆优化，时间复杂度 O(V2)

#include<iostream>#include<cstdio>#include<climits>int head[105],vet[10005],w[10005],next[10005],num;int dist[105];bool f[105];void addedge(int x,int y,int c) {    num++;    vet[num]=y;    w[num]=c;    next[num]=head[x];    head[x]=num;}int main() {    int n,m,ans=0;    std::cin >> n >> m;    for (int i=1;i<=m;i++) {        int x,y,c;        scanf("%d%d%d",&x,&y,&c);        addedge(x,y,c);        addedge(y,x,c);    }    for (int i=1;i<=n;i++) dist[i]=INT_MAX;    dist[1]=0;    for (int i=1;i<=n;i++) {        int mindist=INT_MAX,p=0;        for (int j=1;j<=n;j++)            if (!f[j] && mindist>dist[j]) {                mindist=dist[j];                p=j;            }        f[p]=true;        ans+=dist[p];        for (int e=head[p];e!=0;e=next[e])            if (!f[vet[e]])                dist[vet[e]]=std::min(dist[vet[e]],w[e]);    }    std::cout << ans;    return 0;}

堆优化，时间复杂度 O(ElogE)
priority_queue实现：

#include<iostream>#include<cstdio>#include<climits>#include<queue>struct node {    int p,weight;    bool operator<(const node &a)const {        return weight>a.weight;    }};int head[100005],vet[200005],w[200005],next[200005],num;int dist[100000];bool f[100005];void addedge(int x,int y,int c) {    num++;    vet[num]=y;    w[num]=c;    next[num]=head[x];    head[x]=num;}int main() {    int n,m;    int ans=0;    std::cin >> n >> m;    for (int i=1;i<=m;i++) {        int x,y,c;        scanf("%d%d%d",&x,&y,&c);        addedge(x,y,c);        addedge(y,x,c);    }    for (int i=1;i<=n;i++) dist[i]=INT_MAX;    dist[1]=0;    std::priority_queue<node> qu;    node newnode;    newnode.weight=0;    newnode.p=1;    while (!qu.empty()) qu.pop();    qu.push(newnode);    for (int i=1;i<=n;i++) {        while (f[qu.top().p]) qu.pop();        newnode=qu.top();        qu.pop();        f[newnode.p]=true;        ans+=newnode.weight;        for (int e=head[newnode.p];e!=0;e=next[e])            if (!f[vet[e]] && dist[vet[e]]>w[e]) {                newnode.p=vet[e];                newnode.weight=w[e];                dist[vet[e]]=w[e];                qu.push(newnode);            }    }    std::cout << ans;    return 0;}

反向映射堆：

#include<iostream>#include<cstdio>#include<climits>struct node {    int p,weight;    bool operator<(const node &a)const {        return weight<a.weight;    }};int head[100005],vet[200005],w[200005],next[200005],num;int dist[100005],hsize,po[100005];node heap[200005];bool f[100005];void addedge(int x,int y,int c) {    num++;    vet[num]=y;    w[num]=c;    next[num]=head[x];    head[x]=num;}void sw(int p1,int p2) {    po[heap[p1].p]=p2;    po[heap[p2].p]=p1;    node tmp=heap[p1];    heap[p1]=heap[p2];    heap[p2]=tmp;}void Up(int p) {    while (p>1) {        if (heap[p]<heap[p/2]) {            sw(p,p/2);            p/=2;        }        else break;    }}void Down(int p) {    while (p*2<=hsize) {        int Min=p;        if (heap[p*2]<heap[Min]) Min=p*2;        if (p*2<hsize && heap[p*2+1]<heap[Min]) Min=p*2+1;        if (Min != p) {            sw(p,Min);            p=Min;        }        else break;    }}void decrease_key(int t,int v) {    heap[t].weight=v;    Up(t);}void heap_push(node res) {    heap[++hsize]=res;    po[res.p]=hsize;    Up(hsize);}node heap_pop() {    sw(1,hsize);    hsize--;    Down(1);    po[heap[hsize+1].p]=0;    return heap[hsize+1];}int main() {    int n,m;    int ans=0;    std::cin >> n >> m;    for (int i=1;i<=m;i++) {        int x,y,c;        scanf("%d%d%d",&x,&y,&c);        addedge(x,y,c);        addedge(y,x,c);    }    for (int i=1;i<=n;i++) dist[i]=INT_MAX;    dist[1]=0;    heap_push((node){1,0});    for (int i=1;i<=n;i++) {        node newnode=heap_pop();        f[newnode.p]=true;        ans+=newnode.weight;        for (int e=head[newnode.p];e!=0;e=next[e])            if (!f[vet[e]] && dist[vet[e]]>w[e]) {                if (po[vet[e]]==0) heap_push((node){vet[e],w[e]});                decrease_key(po[vet[e]],w[e]);                dist[vet[e]]=w[e];            }    }    std::cout << ans;    return 0;}

阅读全文

0 0