9.5题解

总分201，rank3
T2图上的简单题，但调了好久，T3暴力分很足，st表加减枝91，T1嘛，卡读题啊，QAQ……
先说坑爹的T1:
先是没看见每种喜悦值只能获得一次，改的时候又发现一次只可以买一个，233
状压每个状态表示每种物品是否被买，转移时可能转移到自己或新的状态，导一下式子倒推就好了。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<ctime>using namespace std;int n,bit[21];long long all,v[21];double p[21],pp,pll,now,f[1<<21];int main(){    bit[0]=1;for(int i=1;i<=20;i++)bit[i]=bit[i-1]<<1;    scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%lf%lld",&p[i],&v[i]);        all+=v[i];        pll+=p[i];    }    f[bit[n]-1]=0;    for(int i=bit[n]-2;i>=0;i--){        pp=0,now=0;        for(int j=1;j<=n;j++){            if(!(i&bit[j-1])){                now+=p[j]*f[i|bit[j-1]];                pp+=p[j];            }        }        f[i]=(now+1.0)/(1.0*pp);    }    printf("%lld\n%0.3lf\n",all,f[0]);}

T2，就是缩点，然后贪心倒着找最小花费

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#define N 100050#define M 200050using namespace std;int head[2*N],e=1;struct edge{    int u,v,w,next;}ed[2*M];void add(int u,int v,int w){    ed[e].u=u;ed[e].v=v;ed[e].w=w;    ed[e].next=head[u];head[u]=e++;}int dfn[N],low[N],top,q[N],tot,id[N];bool bo[N];void tarjan(int x){    dfn[x]=low[x]=++top;    q[top]=x; bo[x]=1;    for(int i=head[x];i;i=ed[i].next){        int v=ed[i].v;        if(!dfn[v]){            tarjan(v);            low[x]=min(low[x],low[v]);        }        else if(bo[v])            low[x]=min(low[x],dfn[v]);    }    if(dfn[x]==low[x]){        int y;tot++;        do{            y=q[top--];            id[y]=tot;            bo[y]=0;        }while(y!=x);    }}int n,m,ans,out[2*N];bool vis[2*N];queue<int> qu;void init(){    e=1;ans=0;top=0;tot=n;    memset(head,0,sizeof head);    memset(dfn,0,sizeof dfn);    memset(vis,0,sizeof vis);    memset(bo,0,sizeof bo);    memset(out,0,sizeof out);}int main(){    while(scanf("%d%d",&n,&m)==2&&!(n==0&&m==0)){        init();        int u,v,w;        for(int i=1;i<=m;i++){            scanf("%d%d%d",&u,&v,&w);            u++;v++;            add(u,v,w);        }        for(int i=1;i<=n;i++)            if(!dfn[i])                tarjan(i);        for(int i=1;i<=m;i++){            int u=ed[i].u,v=ed[i].v;            if(id[u]!=id[v]){                add(id[v],id[u],ed[i].w);                out[id[u]]++;            }        }        for(int i=n+1;i<=tot;i++)            if(!out[i]){                qu.push(i);                vis[i]=1;            }        while(!qu.empty()){            int now=qu.front();qu.pop();            int minn=0x7fffffff,nxt=-1;            for(int i=head[now];i;i=ed[i].next){                int v=ed[i].v;                 if(vis[v])continue;                if(ed[i].w<minn){minn=ed[i].w;nxt=v;}            }            if(nxt==-1)break;            for(int i=head[now];i;i=ed[i].next){                if(vis[v])continue;                out[ed[i].v]--;                if(!out[ed[i].v]){                    qu.push(ed[i].v);                    vis[ed[i].v]=1;                }            }            ans+=minn;        }        printf("%d\n",ans);    }    return 0;}

T3,考试暴力水了91
暴力

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#define N 50050using namespace std;int n,a[N],maxn[N][20],minn[N][20],ans;void init(){    for(int i=1;i<=n;i++)maxn[i][0]=a[i];    for(int i=0;(1<<(i+1))<=n;i++)        for(int j=1;(j+(1<<(i+1)))-1<=n;j++)            maxn[j][i+1]=max(maxn[j][i],maxn[j+(1<<i)][i]);    for(int i=1;i<=n;i++)minn[i][0]=a[i];    for(int i=0;(1<<(i+1))<=n;i++)        for(int j=1;(j+(1<<(i+1)))-1<=n;j++)            minn[j][i+1]=min(minn[j][i],minn[j+(1<<i)][i]);}int work(int x,int y){    int k=0;    while((1<<(k+1))<y-x+1)k++;    int dd=max(maxn[x][k],maxn[y-(1<<k)+1][k]);    int xx=min(minn[x][k],minn[y-(1<<k)+1][k]);    return dd-xx;}int main(){    scanf("%d",&n);    int x,y;    for(int i=1;i<=n;i++){        scanf("%d%d",&x,&y);        a[x]=y;    }    init();    for(int i=1;i<=n;i++){        for(int j=i;j<=n;j++){            int de=work(i,j);            if(de<=j-i)ans++;            else j=i+de-1;        }    }    printf("%d\n",ans);}

正解分治，work(l,r)=work(l,mid)+work(mid+1,r)+跨过mid的，怎么算呢，分为四种情况搞，大小|，|大小,大|小，小|大。
根据max-min=r-l瞎搞就好了。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#define N 300005using namespace std;int maxl[N],maxr[N],minl[N],minr[N],b[4*N],a[N],n;int work(int l,int r){    if(l==r)return 1;    long long ans=0;    int mid=(l+r)>>1;    maxl[mid]=minl[mid]=a[mid];    for(int i=mid-1;i>=l;i--){        maxl[i]=max(maxl[i+1],a[i]);        minl[i]=min(minl[i+1],a[i]);    }    maxr[mid+1]=minr[mid+1]=a[mid+1];    for(int i=mid+2;i<=r;i++){        maxr[i]=max(maxr[i-1],a[i]);        minr[i]=min(minr[i-1],a[i]);    }    for(int i=mid;i>=l;i--){        int d=i+maxl[i]-minl[i];        if(d<=mid||d>r)continue;        if(minr[d]>minl[i]&&maxr[d]<maxl[i])ans++;    }    for(int i=mid+1;i<=r;i++){        int d=i-maxr[i]+minr[i];        if(d>mid||d<l)continue;        if(minl[d]>minr[i]&&maxl[d]<maxr[i])ans++;    }    int r1=mid+1,r2=mid;    for(int i=mid;i>=l;i--){        while(minr[r2+1]>minl[i]&&r2<r){            r2++;            b[maxr[r2]-r2+2*N]++;        }        while(maxr[r1]<maxl[i]&&r1<=r){            b[maxr[r1]-r1+2*N]--;            r1++;        }        if(r1>r)break;        if(r1<=r2)ans+=b[minl[i]-i+2*N];    }    for(int i=mid+1;i<=r;i++)b[maxr[i]-i+2*N]=0;    int l1=mid,l2=mid+1;    for(int i=mid+1;i<=r;i++){        while(minl[l2-1]>minr[i]&&l2>l){            l2--;            b[maxl[l2]+l2+2*N]++;        }        while(maxl[l1]<maxr[i]&&l1>=l){            b[maxl[l1]+l1+2*N]--;            l1--;        }        if(l1<l)break;        if(l1>=l2)ans+=b[minr[i]+i+2*N];    }    for(int i=l;i<=mid;i++)b[maxl[i]+i+2*N]=0;    return ans+work(l,mid)+work(mid+1,r);}int main(){    scanf("%d",&n);    int x,y;    for(int i=1;i<=n;i++){        scanf("%d%d",&x,&y);        a[x]=y;    }    printf("%d\n",work(1,n));    return 0;}