usaco /the first wave

bzoj1572:贪心。先按时间顺序排序,然后用优先队列,如果时间不矛盾直接插入,否则判断队列中w最小的元素是否替换掉。(没用llWA了一次

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--) 
#define ll long long
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=100005;
struct Node{
	int num,w;
	bool operator<(const Node&rhs)const{
	  return num<rhs.num;}
};
Node Nodes[nmax];
struct node{
	int num,w;
	bool operator<(const node&rhs)const{
	  return w>rhs.w;}
};
node nodes[nmax];
priority_queue<node>q;
int main(){
	int n=read();
	REP(i,1,n) Nodes[i].num=read(),Nodes[i].w=read();
	sort(Nodes+1,Nodes+n+1);
	REP(i,1,n) nodes[i].num=Nodes[i].num,nodes[i].w=Nodes[i].w;
	REP(i,1,n){
		node o=nodes[i];
		if(o.num>q.size()) {
			q.push(o);continue;
		}
		node tp=q.top();
		if(tp.w<o.w) q.pop(),q.push(o);
	}
	ll ans=0;
	while(!q.empty()) ans+=q.top().w,q.pop();
	printf("%lld\n",ans);
	return 0;
}

bzoj1574:贪心,将不能到达的点与相连的点集删除,然后dfs。(两个数组名相同RE了一次

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
#define op() clr(head,0);pt=edges;
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=30005;
const int maxn=200005;
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt,*head[nmax];
bool vis[nmax],V[nmax];
void adde(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->next=head[v];head[v]=pt++;
}
void dfs(int x){
	V[x]=true;
	qwq(x) if(vis[o->to]&&!V[o->to]) dfs(o->to);
}
int main(){
	op();clr(vis,true);clr(V,false);
	int N=read(),M=read(),P=read(),u,v;
	REP(i,1,M) u=read(),v=read(),adde(u,v);
	REP(i,1,P) {
		u=read();vis[u]=false;
		qwq(u) vis[o->to]=false;
    }
	dfs(1);
	int ans=N;
	REP(i,1,N) if(V[i]) ans--;
	printf("%d\n",ans);
	return 0;
}

bzoj1576:由于最短路唯一。所以有最短路径树。未在树中的边u,v对于lca(u,v)以下的点,有dist[x]=dist[u]+dist[v]+val[u,v]-dist[x](与dist[x]无关。可以将dist[u]+dist[v]+val[u,v]排序,依次更新。由于先更新的必定更优,于是可以用并查集压缩路径。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=100005;
const int maxn=400005;
const int inf=0x7f7f7f7f;

struct edge{
	int to,dist;edge *next;
};
edge edges[maxn],*pt,*head[nmax];
int d[nmax],dep[nmax],fa[nmax],ans[nmax];
void adde(int u,int v,int d){
	pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;
	pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++;
}

struct node{
	int x,d;
	node(int x,int d):x(x),d(d){}
	bool operator<(const node&rhs) const{
	  return d>rhs.d;}
};
priority_queue<node>q;
void dijkstra(){
	clr(d,0x7f);d[1]=0;dep[1]=0;
	q.push(node(1,0));
	while(!q.empty()){
		node tmp=q.top();q.pop();
		if(d[tmp.x]!=tmp.d) continue;
		qwq(tmp.x) if(d[o->to]>d[tmp.x]+o->dist){
			d[o->to]=d[tmp.x]+o->dist;
			dep[o->to]=dep[tmp.x]+1;fa[o->to]=tmp.x;
			q.push(node(o->to,d[o->to]));
		}
	}
}

struct zc{
	int from,to,dist;
	bool operator<(const zc&rhs) const{
	  return dist<rhs.dist;}
};
zc zcs[maxn];

void getdep(int x){
	
}
int update(int u,int v,int val){
	if(u==v) return u;
	if(dep[u]<dep[v]) swap(u,v);
	if(ans[u]==-1) ans[u]=val-d[u];
	return fa[u]=update(fa[u],v,val);
}

int main(){
	clr(head,0);pt=edges;
	int N=read(),M=read(),u,v,dd;
	REP(i,1,M) {
		u=read(),v=read(),dd=read(),adde(u,v,dd);
		zc &oo=zcs[i];oo.from=u,oo.to=v,oo.dist=dd;
	}
	dijkstra();
	REP(i,1,N) printf("%d:%d\n",i,dep[i]);
	
	int cnt=0;
	REP(i,1,M) {
		zc &oo=zcs[i];
		if(d[oo.from]==d[oo.to]+oo.dist||d[oo.to]==d[oo.from]+oo.dist) continue;
		zc &ee=zcs[++cnt];
		ee.from=oo.from,ee.to=oo.to,ee.dist=d[oo.from]+d[oo.to]+oo.dist;
	}
	sort(zcs+1,zcs+cnt+1);
	
	clr(ans,-1);
	REP(i,1,cnt) update(zcs[i].from,zcs[i].to,zcs[i].dist);
	REP(i,2,N) printf("%d\n",ans[i]);
	return 0;
}

bzoj1585:建图后最小割就可以了。(由于没有处理第一个点的情况WA了一次

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define REP(i,s,t)  for(int i=s;i<=t;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=10005;
const int maxn=200005;
const int inf=0x7f7f7f7f;
struct edge{
	int to,cap;edge *next,*rev;
};
edge edges[maxn],*pt,*head[nmax],*cur[nmax],*p[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d){
	add(u,v,d);add(v,u,0);head[u]->rev=head[v];head[v]->rev=head[u];
}

int cnt[nmax],h[nmax];
int maxflow(int s,int t,int n){
	clr(cnt,0);cnt[0]=n;clr(h,0);
	int flow=0,a=inf,x=s;edge *e;
	while(h[s]<n){
		for(e=cur[x];e;e=e->next)  if(e->cap>0&&h[x]==h[e->to]+1) break;
		if(e){
			a=min(a,e->cap);cur[x]=p[e->to]=e;x=e->to;
			if(x==t){
				while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;
				flow+=a,a=inf;
			}
		}else{
			if(!--cnt[h[x]]) break;
			h[x]=n;
			for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1) h[x]=h[e->to]+1,cur[x]=e;
			cnt[h[x]]++;
			if(x!=s) x=p[x]->rev->to;
		}
	}
	return flow;
}

bool vis[nmax];
int main(){
	clr(head,0);pt=edges;clr(vis,false);
	int N=read(),M=read(),P=read(),s=0,t=N+N+1;
	adde(s,1,inf);
	REP(i,1,M){
		int u=read(),v=read();
		adde(u+u,v+v-1,inf);adde(v+v,u+u-1,inf);
	}
	REP(i,1,P){
		int u=read();vis[u]=true;
	}
	adde(1,2,inf);
	REP(i,2,N){
		if(vis[i]) adde(i+i-1,i+i,inf),adde(i+i,t,inf);
		else adde(i+i-1,i+i,1);
	}
	printf("%d\n",maxflow(s,t,t+1));
	return 0;
}

bzoj1589:tarjan缩点后乱搞即可。(读入优化x=x*1+c-'0',由于自己出的都是小数据所以没有发现WA了一次。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=200005;
const int maxn=400005;
const int inf=0x7f7f7f7f;
struct edge{
	int to;edge *next;
};
edge edges[maxn],*pt,*head[nmax];
void add(int u,int v){
	pt->to=v;pt->next=head[u];head[u]=pt++;
}

int sccno[nmax],scc[nmax],pre[nmax],scc_cnt,dfs_clock=0;
stack<int>s;
int dfs(int x){
	int lowu=pre[x]=++dfs_clock;
	s.push(x);
	qwq(x){
		int to=o->to;
		if(!pre[to]) lowu=min(lowu,dfs(to));
		else if(!sccno[to]) lowu=min(lowu,pre[to]);
	}
	if(lowu==pre[x]){
		scc_cnt++;scc[scc_cnt]=0;
		while(1){
			int tmp=s.top();s.pop();
			sccno[tmp]=scc_cnt;scc[scc_cnt]++;
			if(tmp==x) break;
		}
	}
	return lowu;
}

int sum[nmax];
int getsum(int x){
	if(sum[x]) return sum[x];
	int ans=scc[x];
	qwq(x) ans+=getsum(o->to);
	return ans;
}

int main(){
	clr(head,0);pt=edges;
	int N=read(),u;
	REP(i,1,N) u=read(),add(i,u);
	
	scc_cnt=N;
	REP(i,1,N) if(!pre[i]) dfs(i);
	REP(i,1,N) qwq(i) if(sccno[i]!=sccno[o->to]) add(sccno[i],sccno[o->to]);
	REP(i,N+1,scc_cnt) sum[i]=getsum(i);
	REP(i,1,N) printf("%d\n",sum[sccno[i]]);
	return 0; 	
}

summary:

1.各种神奇的错误方式。。。

2.自己出的数据也不要太小。。。但也不要自己被自己出的数据绕晕了。。。

posted @ 2016-07-22 11:43  BBChq  阅读(134)  评论(0编辑  收藏  举报