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
%0.3lf
",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 200050
using 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
",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 50050
using 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
",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 300005
using 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
",work(1,n));
    return 0;
}