一些代码在压缩路径同时合并距离

intfind_set(int{if(pre[x]==x)returnelsereturnpre[x]=}voidunion_set(intx,int{intxx=find_set(x);intyy=if(xx!=yy)pre[yy]=}intFind(int{returnintreturnparent[x]=tmp;}intUnion(intu,intv,intintreturn0;}void{for(inti=2;i<N;i++){if(!pt[i])pr[pn++]=i,pt[i]=i;for(intj=0;j<pn&&pr[j]*i<N;j++){pt[pr[j]*i]=pr[j];if(i%pr[j]==0)}}}intsolve(int{intans=1;while(n!=1){intintk=pt[n];//ptwhile{}}return}Primvoid{intintmin_id,now=0;doublemin_edge;doubledist[N];inti,for(i=0;i<n;{fa[i]=-dist[i]=}fori0;in1;i++)//{min_edge=forj0;jnj++)//{if(mm[now][j]>{if(mm[now][j]<{dist[j]=mm[now][j];fa[j]=now;}}}dist[now1.0;//已经加入生成树的点的距离置为-1for(j=0;j<n;j++){ifdist[j]>0&&dist[jmin_edge)//{min_edge=dist[j];min_id=j;}}nowmin_id;//更新新加进来的点sum+=min_edge;}}SPFA邻接链表#include<iostream>#include<queue>usingnamespacestd;constlonglmax constlongedge_maxn=10005; constlongpoint_maxn=105; structnode{/*node边，一个edge代表一条边*/intv; int intnext; //同一起点的下一条边在edge数组中的位置（理解了这个静态邻接表就int //以该点为起点的第一条边在edge数组中的位intn;//点的数量intm;//边的数量queue<int>Q;intdirs[point_maxn];boolvis[point_maxn];voidInit(){intIndex=1;inti,j,flag;intx,y,w;{if(y=={if(w<edge[j].w=w;flag=1;}}if(flag==edge[Index].v=y;edge[Index].w=edge[Index].next //xedgepre[x]=Index++; edge[Index].v=y;edge[Index].w=edge[Index].next=pre[x];pre[x]=Index++;}}voidprint(int{}void{intstart=1;intend=n;{}dirs[start]=vis[start]=1;{inttopQ.front();边的起点vis[top]={inteedge[j].v;if(dirs[e]>edge[j].w+{dirs[e]=edge[j].w+dirs[top];{vis[e]=}}}}}int{while(scanf("%d%d",&n,&m)!=EOF&&(n!=0||{}return}struct{doublejung(juaa,jubb,int{jucc;doubleans;{{{}}}return}{{ans=}oo=}0for(i=0;i<n;for(j=0;j<n;if(a[i][j]==0)continue;for(k=0;k<n;{c[i][k]+=a[i][j]*}}}TarjanvoidInit_Tarjan(intn){for(inti=1;i<=n;i++){dfn[i]=low[i]=Belong[i]=0;}}voidTarjan(intu){intv;} }}}intfor(i=1;i<=n;i++)if(!dfn[i])Tarjan(i);for(i=1;i<=n;i++)for(j=0;j<G[i].size();{k=if(Belong[i]!=}#include<stdio.h>#definemaxint64int64voidMerge(longlonga[],intstart,intmid,int{inti=start,j=mid+1,k=start;while(i<=mid&&j<=end){if(a[i]<={}{}}

b[k++]=count+=j-k;b[k++]=a[j++];while(i<={b[k++]=}while(j<={b[k++]=}for(inti=start;i<=end;{a[i]=}}voidMergeSort(longlonga[],intstart,int{if(start<{intmid=(start+end)/2;}}intmain(intargc,charconst{int64{count0;//记录逆序数个数for(inti=0;i<m;i++){}}return}constintmod vector<int64> int64 //初始化的时候要for(i=0;i< ;i++)ha[i].clear();voidinsert(int64x){intp=abs(x)%mod;}intfind(int64{intp=abs(x)%for(inti=0;i<ha[p].size();i++)if(ha[p][i]==x)return1;return0;}while{inttmp=word[i]-if(now->next[tmpNULL)//{}{}

trie*nextnewtrie();//newnow->next[tmp]=nownow->next[tmp];//dfs}}intsearch(trie*root,char*{trie*now=root;inti=0;while{inttmp=word[i]-if(now->next[tmpNULL)return0;//如果没有下一个那么就没有要找的前缀now=now->next[tmp];}returnnow->count;//return0}int{trie*rootnewtrie();//}struct{intx,}nn[510],e[510],intcmp(nodea,node{if(a.x==b.x)returna.y<b.y;returna.x<b.x;}intcross(nodea,nodeb,nodec)//{return(a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-}{sort(e,e+n,intm0,i,jk;////mi，则向量(m-2,i)必定要在(m-2,m-1)的逆时针方向m-1点不在凸包上。for(i=0;i<n;{while(m>1&&cross(res[m-1],e[i],res[m-2])<=0)m--;res[m++]=e[i];}k=m;//求得上凸包forin2;i0;i--){while(m>k&&cross(res[m-1],e[i],res[m-2])<=0)m--;res[m++]=e[i];}ifn>1)m--;//returnm;（处理完后凸包上的点的个数}for(i=0;i<marki;{for(j=0;j<len;{if(cross(res[j],e[i],res[(j1len])0)//e[i]}}constintMAX_M=201;constintMAX_N=201;vector<int>g[MAX_M+MAX_N];//这里最大一定要设置两个值得和intmatch[MAX_MMAX_N];//1~N和N+1~N+M两部分intuse[MAX_M+MAX_N];intv,n,voidadd_edge(intu,int{g[u].push_backg[v].push_back(u);//}intdfs(intx)//dfs{inti,j;use[x]=1;for(i=0;i<g[x].size();{intu=g[x][i],w=if(w<0||!use[w]&&dfs{match[x]=u;match[u]=x;return1;}}return}inthungary{

intdfs(intx)//{inti,for(i=0;i<g[x].size();{intu=g[x][i],w=match[u];if(use[u])continue;use[u]=if(w<0||{match[u]=x;return1;}}return}intres=memset(match,1,sizeof(match));//单方匹配去掉这个for(v=1;v<=n+m;v++){if(match[v]==-{memset(use,0,sizeof(use));if(dfs(v))res++;}}return}intmain()中要注意的地方：fori0;iMAX_MMAX_N;i++)g[i].clearadd_edge(n+u,i);constintMAX_N100010;/mnconstintMAX_M11;/nintmm[MAX_N][MAX_M],match[MAX_M][MAX_N],use[MAX_M],limit[MAX_M],has[MAX_M];intv,n,m;match[MAX_M][MAX_Nmimatch[MAX_M][MAX_Nmiuse[MAX_M]nmlimit[MAX_M]mnhas[MAX_M]mn{inti,for(i=0;i<m;{if(mm[x][i]&&{use[i]=if(has[i]<{match[i][has[i]++]=x;return1;}for(j=0;j<limit[i];{if{match[i][j]=x;return1;}}}}return}inthungary()inthungary(){intres=memset(has,0,memset(match,-1,sizeof(match));for(v=0;v<n;v++){memset(use,0,sizeof(use));if(dfs(v))res++;}return}{memset(has,0,memset(match,-1,sizeof(match));for(v=0;v<n;v++){memset(use,0,sizeof(use));if(!dfs(v)){return}}return}intdfs(int{for(inti=0;i<bg[x].size();{intv=if{vis[v]=if(match[v]==-1||{match[v]=x;return1;}}}return}void{intcnt=memset(match,-1,sizeof(match));for(inti=1;i<=CNT;i++){memset(vis,0,sizeof(vis));if(dfs(i))cnt++;}printf("%d\n",CNT-}最大流，Dinicvoidbfs(int{memset(level,-1,queue<int>que;level[st]=0;que.push(s);while(!que.empty()){intv=que.front();que.pop();for(inti=0;i<n+2;i++){if(flow_map[v][i]>0&&level[i]<{level[i]=level[v]+1;que.push}}}}intdfs(intvintuintf)//u{if(v==u)returnfor(inti=0;i<n+2;{if(flow_map[v][i]>0&&level[v]<{intd=dfs(i,u,min(f,flow_map[v][i]));if(d>0){flow_map[v][i]-=d;flow_map[i][v]+=d;return}}}return}intmax_flowintst,inted)//Dinic{intflow=0;for(;;){bfsif(level[ed]<0)returnflow;intf;while((f=dfs(st,ed,inf))>0)flow+=}}最大流，Dinicintlevel[边最大];intiter[边最大];structedge{intto,cap,voidadd_edge(intfrom,intto,int{edgee;e.to=to;e.cap=e.rev=G[to].sizeG[from].push_back(e);e.to=from;e.cap=e.rev=G[from].size()-}voidbfs(int{memset(level,-1,sizeof(level));queue<int>que;level[s]=0;que.push(s);while{intv=que.front();que.pop();for(inti=0;i<G[v].size();i++){edge&e=if(e.cap>0&&level[e.to]<{level[e.to]=level[v]+1;que.push(e.to);}}}}intdfs(intv,intt,int{if(v==t)returnfor(int&i=iter[v];i<G[v].size();{edge&e=if(e.cap>0&&level[v]<{intd=dfs(e.to,t,min(f,e.cap));if(d>0){e.cap-=G[e.to][e.rev].cap+=d;returnd;}}}return}intmax_flow(ints,int{intflow=0;for(;;){bfsif(level[t]<0)returnmemset(iter,0,sizeof(iter));;intf;while((f=dfs(s,t,inf))>0){flow+=f;}}KMPvoidgetNxt()nxt[0]=-inti=0,j=-while(i<p.size())if(j==-1||p[i]==p[j])ijnxt[i]=}}}

j=intkmp()inti=0,j=0,ans=0;while(i<s.size()){if(j==-1||s[i]==p[j])ij}j=if(j==ans}return}intn,intindegree[maxn0};//每个点的入边条数vector<int>v[maxn];//邻接链表set<int>intans[maxn];voidreadin(){intscanf("%d%d",&n,&m);for(i=1;i<=n;i++)while(m--)scanf("%d%d",&i,&jv[i].push_back(j);indegree[j]++;}}int{inti,pos=0;for(i=1;i<=n;i++if(indegree[i]==0)s.insert(i);while(!s.empty(){i=*s.begin();s.erase(s.begin()while(!v[i].empty(){if(--indegree[v[i].back()]==0s.insert(v[i].back());v[i].pop_back();}}ifreturn}

returnconstintMAX_N=200010;#definelson(x)((x)<<1)#definerson(x)((x)<<1|1)struct{intl,r,}nn[MAX_N*3];intintbuild_tree(intleft,intright,int{intnn[i].l=left;nn[i].r=right;if(left==right){returnnn[i].maxx=}mid=(left+right)/returnnn[i].maxx=max(build_tree(left,mid,lson(i)),build_tree(mid+1,right,}voidupdate(intcid,intva,int{intmid=(nn[i].l+nn[i].r)/if(nn[i].l=={}{}}

nn[i].maxx=va;if(cid<=mid)update(cid,va,lson(i));elseupdate(cid,va,rson(i));nn[i].maxx=max(nn[lson(i)].maxx,intquery(intleft,intright,int{intif(nn[i].l==left&&nn[i].r==right)returnnn[i].maxx;mid=(nn[i].l+nn[i].r)/2;if(right<=mid)returnquery(left,right,elseif(left>mid)return query(left,right,rson(i));{inta=query(left,mid,intb=query(mid+1,right,rson(i));returnmax(a,b);}}query(l,r,change(cid,va,lazyconstintMAX_N100010;#definelson(x)((x)<<1)#definerson(x)((x)<<1|1)intstruct{intl,r,sum;intlazy;}nn[MAX_N*4];voidpush_up(inti){nn[i].sum=nn[lson(i)].sum+}voidpush_down(int{if{nn[lson(i)].lazy=nn[rson(i)].lazy=nn[lson(i)].sum=(nn[lson(i)].r-nn[lson(i)].l+nn[rson(i)].sum=(nn[rson(i)].r-nn[rson(i)].l+1)*nn[rson(i)].lazy;nn[i].lazy=0;}}voidbuild_tree(intleft,intright,int{intnn[i].l=left;nn[i].r=right;nn[i].lazy0;//lazy0，不单单叶子节点if(left==right){nn[i].sum=1;}mid=(left+right)/2;build_tree(left,mid,lson(i));build_tree(mid+1,right,rson(i));}voidupdate(intleft,intright,intva,int{intmid=(nn[i].l+nn[i].r)/if{nn[i].lazy=nn[i].sum=(nn[i].r-nn[i].l+1)*va;}if(right<=mid)update(left,right,va,lson(i));elseif(left>mid)update(left,right,va,rson(i));{update(left,mid,va,lson(i));update(mid+1,right,va,rson(i));}}voidquery(intleft,intright,int{intif(nn[i].l==left&&nn[i].r=={ans+=nn[i].sum;}mid=(nn[i].l+nn[i].r)/if(right<=mid)query(left,right,elseif(left>mid)query(left,right,rson(i));{query(left,mid,lson(i));query(mid+1,right,rson(i));}}HDU48932014（可作为线段树模板）n0的k kl lrlr 改变从l到r所有元素，都为原元素最近的Fibonacci数，差值相等时取较小值处理的方案就是提前对每一段的区间和都找到相应的Fibonacci#include<cstdio>#include<cmath>usingnamespacestd;#defineN100010structnodeintL,longlongsum,intflag;//1vsumsumlazy}tree[N<<2];longlonglonglongffib(longlongval){longlongx=0,y=1;intfor(i=0;i<100;++i){y=x+y;x=y-if(y>=}if(fabs(y-val)<fabs(x-val))returny;return}voidPushUP(intp)if(tree[p].L==tree[p].R)tree[p].sum=tree[p<<1].sum+tree[p<<1|1].sum;tree[p].vsum=tree[p<<1].vsum+tree[p<<1|1].vsum;}voidPushDown(intp)if(tree[p].flag&&tree[p].Ltree[p].R是叶子节点就不用往下更了tree[p].sum=tree[p].vsum;tree[p].flag=0;}if(tree[p].flag)tree[p1].flagtree[p1|1].flag1;//儿子要继续往下更tree[p<<1].sum=tree[p<<1].vsum;tree[p<<1|1].sum=tree[p<<1|1].vsum;tree[p].flag0;//0}}voidbuild(intL,intR,intp){tree[p].L=L;tree[p].R=R;tree[p].flag=0;if(L==R){tree[p].sum0;//0tree[p].vsum=1;//0fib1fibsum1}intmid=(L+R)>>1;build(L,mid,p<<1);build(mid+1,R,p<<1|1);PushUP(p);//sum和vsumvsum，sum}voidadd(intpos,intval,intp)if(tree[p].Ltree[p].R)确定要叶子节点，也就是要加的那个点时tree[p].sum+=val;tree[p].vsumffib(tree[p].sum);//fibsumtree[p].flag=0;}intmid=(tree[p].L+tree[p].R)>>1;if(mid>=pos)add(pos,val,p<<1);elseadd(pos,val,p<<1|1);}voidupdate(intL,intR,intp)if(L<=tree[p].L&&R>=tree[p].R)tree[p].flag1;//1tree[p].sum=tree[p].vsum;}PushDown(p);//三个函数都要有PushDown和PushUpflagupdate的时候1intmid=(tree[p].L+tree[p].R)>>1;if(mid>=R)update(L,R,p<<1);elseif(mid+1<=L)update(L,R,p<<1|1);else{update(L,mid,p<<1);update(mid+1,R,p<<1|1);}voidquery(intL,intR,intp)if(L<=tree[p].L&&R>=tree[p].R){ans+=tree[p].sum;}intmid=(tree[p].L+tree[p].R)>>1;if(mid>=R)query(L,R,p<<1);elseif(mid+1<=L)query(L,R,p<<1|1);else{query(L,mid,p<<query(mid+1,R,p<<1|}intmain()//freopen("in.txt","r",stdin);intn,m,i;intop,x,while(scanf("%d%d",&n,&m)!=EOF){build(1,n,1);for(i=0;i<m;++i)scanf("%d%d%d",&op,&x,&y);if(op==1){add(x,y,}elseif(op==2){ans=0;query(x,y,printf("%I64d\n", //}elseif(op==3){update(x,y,1);}}}return}HDU49712014多校赛一共有n个任务，m个技术，完成一个任务可一些钱，学个#include<cstdio>#include<cstring>#include<vector>#include<algorithm>#definesz1000#defineinf0x3fffffffusingnamespacestd;intn,m;inta[sz],b[sz];vector<int>p[sz];vector<int>g[sz],intlow[sz],dfn[sz],stack[sz],instack[sz],belong[sz],cost[sz];intdeep,top,kinds;voidtarjan(ints)//tarjan模{low[s]=dfn[s]=deep++;stack[++top]=s;instack[s]=1;for(inti=0;i<g[s].size();{intt=if(dfn[t]==-{tarjanlow[s]=min(low[t],}elseif{low[s]=min(dfn[t],}}if(low[s]=={intintct=0;{t=stack[top--belong[t]=kinds;//每个属于哪个缩点后的点instack[t]=0;ct+=b[t];}while(t!=cost[kinds]ct;//cost记录缩点后的点集中包含的几个技术的花费和}}structnode{ints,t,v,nxt;}e[sz*inthd[sz],dis[sz],cur[sz],que[sz],cnt,S,T;voidinsert(ints,intt,intv){e[cnt].s=s,e[cnt].t=t,e[cnt].v=v,e[cnt].nxt=hd[s],hd[s]=e[cnt].s=t,e[cnt].t=s,e[cnt].v=0,e[cnt].nxt=hd[t],hd[t]=}boolbfs(ints,intt,intn){inthead=0,tail=0,i;memset(dis,-1,sizeof(dis[0])*(n+1)),dis[s]=0;que[tail++]=s;whilefor(i=hd[que[head++]];i!=-1;i=if(e[i].v>0&&dis[e[i].t]==-dis[e[i].t]=dis[e[i].s]+1;if(e[i].t==t)return1;que[tail++]=e[i].t;}}return}intdinic(ints,intt,intn){//最大流模板,这个无视，暂时还只会套模板，有点看不懂inti,mf=0,tp;while(bfs(s,t,for(i=0;i<n;i++)cur[i]=hd[i];intu=s,tail=0;while(cur[s]!=-if(u!=t&&cur[u]!=-1&&e[cur[u]].v>0&&dis[u]+1==dis[e[cur[u]].t]){que[tail++]=cur[u];u=e[cur[u]].t;}elseif(u==for(tp=inf,i=tail-1;i>=0;i--)tp=min(tp,e[que[i]].v);for(mf+=tp,i=tail-1;i>=0;i--){e[que[i]].v-=tp;e[que[i]^1].v+=tp;if(e[que[i]].v==0)tail=i;}u=}elsewhile(u!=s&&cur[u]==-1)u=e[que[--tail]].s;cur[u]=e[cur[u]].nxt;}}}return}voidmake_map{for(inti=0;i<kinds+n;i++)ng[i].clear();for(inti=0;i<m;for(intj=0;j<g[i].size();{intt=if(belong[i]!={ng[belong[i]].push_back}}for(inti0;i<n;i++)//下标[0,kinds)是技术的缩点,[kinds,kinds+n)是项目的{intid=i+for(intj=0;j<p[i].size();ng[id].push_back//项目和技术之间建边，belong[p[i][j]]技术所属缩点后点}for(inti=0;i<kinds+n;{sort(ng[i].begin(),ng[i].endng[i].resize(unique(ng[i].begin(),ng[i].end())-ng[i].begin//一个项目可能用到的多个技术缩点后在同一个点集中，所以取}memset(hd,-1,sizeof(hd));cnt=0;S=kinds+T=S+for(inti0;i<n;i++)//源点s往所有的项目建流量为收益的边insert(S,i+kinds,a[i]);for(inti=0;i<n+kinds;for(intj0;j<ng[i].sizej++)//中间的点的流量全是infinsert(i,ng[i][j],inf);for(inti0;i<kinds;i++)//所有技术往汇点t建权值为花费的insert(i,T,cost[i]);}intmain{intscanf("%d",for(intcas=1;cas<=cs;{scanf("%d%d",&n,&m);intsum=0;for(inti=0;i<n;scanf("%d",&a[i]),sum+=a[i];for(inti=0;i<m;i++)scanf("%d",&b[i]);intnum,c;for(inti=0;i<n;{scanf("%d",p[i].clearfor(intj=0;j<num;{scanf("%d",p[i].push_back(c);//p用来记录每个项目需要哪些技}}for(inti=0;i<m;{g[i].clearfor(intj=0;j<m;{scanf("%d",&c);if(c)g[i].push_back(j);//有向图记录各个技术之间的先后关}}memset(dfn,-1,sizeof(dfn));top=deep=kinds=0;for(inti=0;i<m;i++)if(dfn[i]==-1)tarjan(i);//技术成环，所以缩点make_map最大闭合权图，建图是个关键intret=sum-dinic(S,T,T+1);printf("Case#%d:%d\n",cas,ret);}return}等于k，则这样称为法。给定n，k，L，求方案数。11（10表示不能）第2为2d[1<<n]DPnk位为1的就可以加上方法数。状态表示111：1能得到，2能得到，3101：1能得到，2得不到，3#include<iostream>#defineLLlonglong#defineMOD#definedebug(x)printf(#x"=%d\n",x);usingnamespacestd;intint{inttt,n,k,L;{intw=(1<<k)-1;//w1inttot=0;if(L>k)//totk能取的数l超过(k,L]这部分用不上，{}while(n--)//n{{intLL(k,L]的个数，k>l0{intsta=i|(1<<(j-加上一个值就是下一个状态能达到的那些数。(1<<(j-1))是从0开始加j因为0位没法位移，1位移然后或上去。最后&w是截掉超过最大状态的那一部分}}}intif((i>>(k-1))&1)//k{}}}n41#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>usingnamespacestd;typedefint64LL;constLLNUM=10;//运算次数，Miller_Rabin2^(-NUM);LLt,f[100];LLmul_mod(LLa,LLb,LLn)//a*b%nab{LLs=0;{}return}LLpow_mod(LLa,LLb,LLn)//{LL{}return}boolcheck(LLa,LLn,LLr,LL{LLans,p,i;{if(ans==1&&p!=1&&p!=n-1)returntrue;}if(ans!=1)returntrue;returnfalse;}boolMiller_Rabin(LLn)//Miller_Rabinn{if(n<2)returnfalse;if(n==2)returntrue;LLi,r,s,a;{return}return}LL(LLa,LL{returnb==0?a:}LLPollard_rho(LLn,LLc)//Pollard_rh