刷刷POI，整个人都POI！
BZOJ 1098: [POI2007]办公楼biu 补图联通块直接DFS出来就行了，很容易TLE要用链表及时删点、
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//WNJXYK //while(true) RP++; #include<iostream> #include<cstdio> #include<algorithm> #include<map> #include<set> #include<queue> #include<string> #include<cstring> using namespace std; const int Maxn=100010; const int Maxm=2000010; inline int read(){ int x = 0; char ch = getchar(); while (ch < '0' || ch > '9') ch = getchar(); while (ch >= '0' && ch <= '9'){ x = x * 10 + ch - '0'; ch = getchar(); } return x; } struct Edge{ int v,nxt; Edge(){} Edge(int v0,int n0){ v=v0; nxt=n0; } }; int head[Maxn]; Edge e[Maxm*2]; int nume; int n,m; inline void addEdge(int u,int v){ e[++nume]=Edge(v,head[u]);head[u]=nume; e[++nume]=Edge(u,head[v]);head[v]=nume; } int nxt[Maxn],pre[Maxn]; inline void del(int x){ int tmp=pre[x]; nxt[tmp]=nxt[x]; pre[nxt[x]]=tmp; } bool vis[Maxn]; int cnt; int siz[Maxn]; bool noEdge[Maxn]; int que[Maxn]; int l,r; int i; void bfs(int src){ que[0]=src; vis[src]=true; for (l = r = 0; l <= r; ++l){ int x=que[l]; //printf("%d\n",x); siz[cnt]++; for (i=head[x];i;i=e[i].nxt) noEdge[e[i].v]=true; for (i=nxt[0];i<=n;i=nxt[i]){ if (!vis[i] && !noEdge[i]){ del(i); vis[i]=true; que[++r]=i; } } for (i=head[x];i;i=e[i].nxt) noEdge[e[i].v]=false; } } inline void input(){ freopen("1098.in","r",stdin); //freopen("1098.out","w",stdout); } int main(){ //input(); n=read();m=read(); int x,y; for (i=1;i<=m;i++){ x=read(),y=read(); addEdge(x,y); } for (i=0;i<=n;i++){ nxt[i]=i+1; pre[i+1]=i; } for (i=nxt[0];i<=n;i=nxt[i]){ if (!vis[i]){ cnt++; del(i); bfs(i); } } sort(siz+1,siz+cnt+1); printf("%d\n",cnt); for (i=1;i<=cnt;i++){ printf("%d ",siz[i]); } printf("\n"); return 0; } |
BZOJ 1097: [POI2007]旅游景点atr 首先SPFA出前k+1个点的最短路,然后做状压DP. pre […]
Nim游戏
BZOJ 1299: [LLH邀请赛]巧克力棒 对于一个初始局面,如果我们可以把它分成一个P态和一个N态.我们一次头把P态全部取出,对于这个P态我们就是后手了.我们强迫对方取出N态的巧克力,这样对于这个N态我们变成了先手,可以保证必胜了.
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//WNJXYK //while(true) RP++; #include<iostream> #include<cstdio> #include<algorithm> #include<map> #include<set> #include<queue> #include<string> #include<cstring> using namespace std; const int Maxn=20; int n; int num[Maxn]; int sum; bool dfs(int now,int uesd,int sg){ if (sg==0 && uesd>0) return true; if (now>n) return false; return dfs(now+1,uesd,sg)||dfs(now+1,uesd+1,sg^num[now]); } int main(){ for (int i=1;i<=10;i++){ scanf("%d",&n); for (int j=1;j<=n;j++){ scanf("%d",&num[j]); } if (!dfs(1,0,0)) printf("YES\n"); else printf("NO\n"); } return 0; } |
