1、HDU 1102 Constructing Roads 最小生成树
2、总结:
题意:修路,裸题
(1)kruskal
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//kruskal #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b using namespace std; #define LL long long #define INF 0x3f3f3f3f const int N=110; int n,k; int father[N]; struct Eage { int st,en,val; }eage[N*(N+1)/2]; bool cmp(Eage a,Eage b) { return a.val<b.val; } int findn(int x) //即并查集的查找 { int t=x; while(x!=father[x]){ x=father[x]; } int p; while(t!=x){ p=t; t=father[t]; father[p]=x; } return x; } int kruskal() { int sum=0; sort(eage,eage+k,cmp); for(int i=0;i<k;i++){ int x=findn(eage[i].st); int y=findn(eage[i].en); if(x!=y){ sum+=eage[i].val; x=findn(x); y=findn(y); father[x]=y; } } return sum; } int main() { while(scanf("%d",&n)!=EOF) { int m; k=0; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ scanf("%d",&m); if(i<j){ eage[k].st=i; //st边起点,en边终点,k为边的条数 eage[k].en=j; eage[k++].val=m; } } } for(int i=1;i<=n;i++){ father[i]=i; } int q,a,b; scanf("%d",&q); while(q--){ scanf("%d%d",&a,&b); a=findn(a); b=findn(b); father[a]=findn(b); //把a与b连通 } int ans=kruskal(); printf("%d ",ans); } return 0; }
(2)prim
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//prim #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b using namespace std; #define LL long long #define INF 0x3f3f3f3f const int N=110; int n; int mapn[N][N]; int visit[N],dis[N]; //dis存储每个点到所选结点的最短距离 int prim() { int sum=0; memset(visit,0,sizeof(visit)); visit[1]=1; for(int i=1;i<=n;i++){ //先选第一个结点 dis[i]=mapn[1][i]; } for(int i=1;i<n;i++) //注,<不是<= { int next,minn=INF; for(int j=1;j<=n;j++){ //找到下一个最近的结点 if(!visit[j]&&minn>dis[j]){ next=j; minn=dis[j]; } } sum+=minn; visit[next]=1; for(int j=1;j<=n;j++){ //更新dis if(!visit[j]&&dis[j]>mapn[next][j]){ dis[j]=mapn[next][j]; } } } return sum; } int main() { int q,a,b; while(scanf("%d",&n)!=EOF) { memset(mapn,INF,sizeof(mapn)); for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ scanf("%d",&mapn[i][j]); } } scanf("%d",&q); while(q--){ scanf("%d%d",&a,&b); mapn[a][b]=mapn[b][a]=0; } int ans=prim(); printf("%d ",ans); } return 0; }