标签:har include noi pre stream struct 并且 hnoi2006 iostream
给定一个n个点m条边的图,每条边都有两个等级,分别要c1和c2费用。你需要构造一个最小生成树,并且满足最少有k条一级边。
n,k<=10000,m<=20000
题解:很自然的想到二分一个答案,然后先连一级边,再连二级边,判断能否连通/是否有足够的一级边 就可以啦。
复杂度mlogMAXC
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int read() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘) f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();} return x*f; } int n,m,K,s[10005]; struct edge{ int from,to,w; }c1[20005],c2[20005]; bool cmp(edge x,edge y){return x.w<y.w;} int getfa(int x){return s[x]==0?x:s[x]=getfa(s[x]);} bool check(int mid) { memset(s,0,sizeof(s));int num=1,cnum=0,x,y; for(int i=1;i<=m&&c1[i].w<=mid;i++) if((x=getfa(c1[i].from))!=(y=getfa(c1[i].to))) s[x]=y,num++,cnum++; if(cnum<K)return false; for(int i=1;i<=m&&c2[i].w<=mid;i++) if((x=getfa(c2[i].from))!=(y=getfa(c2[i].to))) s[x]=y,num++; return num==n; } int main() { n=read();K=read();m=read(); for(int i=1;i<m;i++) { c1[i].from=c2[i].from=read(); c1[i].to=c2[i].to=read(); c1[i].w=read();c2[i].w=read(); } sort(c1+1,c1+m+1,cmp);sort(c2+1,c2+m+1,cmp); int l=1,r=30000,mid,ans; while(l<=r) { mid=l+r>>1; if(check(mid)) ans=mid,r=mid-1; else l=mid+1; } cout<<ans; return 0; }
标签:har include noi pre stream struct 并且 hnoi2006 iostream
原文地址:http://www.cnblogs.com/FallDream/p/bzoj1196.html