【POI2013】bzoj3426 Tower Defence Game
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只需要每次找到一个没有被覆盖的点选掉就可以了。考虑新方案和原方案,最坏情况是没有选到原方案选的点,而是选到了它旁边的点。但是即使这样,新点覆盖的范围也是只多不少。
注意此题没有spj,需要从小到大枚举,也就是输出字典序最小的解。
#include<cstdio>#include<algorithm>const int maxn=2000010;int fir[maxn],ne[maxn],to[maxn],tag[maxn],vis[maxn],ans[maxn],n,m,k,num;void add(int num,int u,int v){ ne[num]=fir[u]; fir[u]=num; to[num]=v;}int main(){ int u,v; scanf("%d%d%d",&n,&m,&k); for (int i=1;i<=m;i++) { scanf("%d%d",&u,&v); add(i<<1,u,v); add(i<<1|1,v,u); } for (int i=n;i;i--) if (!vis[i]) { vis[i]=1; ans[++num]=i; for (int j=fir[i];j;j=ne[j]) { vis[to[j]]=1; for (int k=fir[to[j]];k;k=ne[k]) vis[to[k]]=1; } } printf("%d\n",num); for (int i=1;i<=num;i++) printf("%d ",ans[i]);}
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