NOIP2017 Day1 小结

T1

结论题

预计得分100

复制代码
#include<stdio.h>
#include<iostream>
using namespace std;
typedef long long ll;
ll a,b;
int main(){
    cin>>a>>b;
    cout<<1ll*a*b-1ll*a-1ll*b;
    return 0;
}
复制代码

T2

细节题

栈模拟

垃圾细节题，写挂了[吐血]

预计得分50

复制代码
#include<stdio.h>
#include<iostream>
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
int T,data;
int n,tot,f,e,ans,cnt,flag,l,r;
char S[2333];
char c;
int vis[2333],keep[2333];
struct node{
    char c;
    char name;
    char l[23];
    char r[23];
}q[23333];
inline int read(){
    char c;while(c=getchar(),c==' '||c=='\n');int data=c-48;
    while(c=getchar(),c>='0'&&c<='9')data=(data<<1)+(data<<3)+c-48;return data;
}
int main(){
    scanf("%d",&T);
    die:;
    while(T--){
        tot=0;
        n=read();
        cin>>S;
        f=0;cnt=ans=0;tot=0;
        for(register int i='a';i<='z';++i)
            vis[i]=0;
        for(register int i=1;S[i]!='\0';++i){
            if(S[i]=='1'&&f==0){
                tot=0;
                break;
            }
            if(S[i]=='^'){
                tot=0;
                f=1;
            }
            if(S[i]>='0'&&S[i]<='9')
                tot=tot*10+S[i]-48;
        }
        e=f=0;
        scanf("\n");
        FOR(i,1,n){    
            scanf("%c ",&q[i].c);
            if(q[i].c=='F'){
                scanf("%c ",&q[i].name);
                scanf("%s ",q[i].l);
                scanf("%s ",q[i].r);
                ++f;
            }
            else
                ++e;
            scanf("\n");
        }
        if(e!=f){
            puts("ERR");
            continue;
        }
        f=0;
        FOR(i,1,n){
            if(q[i].c=='F'){
                ++f;
                keep[f]=i;
                if(vis[q[i].name]){
                    puts("ERR");
                    goto die;
                }
                if(q[i].l[0]!='n'&&q[i].r[0]=='n'&&flag==0)
                    ++cnt;
                if(q[i].l[0]=='n'&&q[i].l[0]!='n')
                    ++flag;
                if(q[i].l[0]!='n'&&q[i].r[0]!='n'){
                    l=0;r=0;
                    for(register int j=0;q[i].l[j]!='\0';++j)
                        l=l*10+q[i].l[j]-48;
                    for(register int j=0;q[i].r[j]!='\0';++j)
                        r=r*10+q[i].r[j]-48;
                    if(l>r)
                        ++flag;
                }
                ans=max(ans,cnt);
                vis[q[i].name]=1;
            }
            else{
                vis[q[keep[f]].name]=0;
                if(q[keep[f]].l[0]!='n'&&q[keep[f]].r[0]=='n'&&flag==0)
                    --cnt;
                if(q[keep[f]].l[0]=='n'&&q[keep[f]].r[0]!='n')
                    --flag;
                if(q[keep[f]].l[0]!='n'&&q[keep[f]].r[0]!='n'){
                    l=0;r=0;
                    for(register int j=0;q[keep[f]].l[j]!='\0';++j)
                        l=l*10+q[keep[f]].l[j]-48;
                    for(register int j=0;q[keep[f]].r[j]!='\0';++j)
                        r=r*10+q[keep[f]].r[j]-48;
                    if(l>r)
                        --flag;
                }
                --f;
                if(f<0){
                    puts("ERR");
                    goto die;
                }
            }
        }
        ans==tot?puts("Yes"):puts("No");
    }
    return 0;
}
复制代码

T3

DP题

真心不会.

只会k=0的做法

跑最短路，然后由近到远进行转移

剩下的点跑A*，好像RE光了

预计得分30

?
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#include<stdio.h>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
#define VIS(now) for(register int e=las[now];e;e=nxt[e])
#define pa pair<int,int>
using namespace std;
typedef long long ll;
priority_queue<pa>heap;
queue<int>q;
const int inf=1<<29;
const int M=400011,N=100011;
int dp[N];
int n,m,k,p,tot,T,ans,f;
int nxt[M],las[N],to[M],w[M],dis[N],vis[N];
int x[M],y[M],z[M];
inline void add(int x,int y,int z){
    nxt[++tot]=las[x];las[x]=tot;
    to[tot]=y;w[tot]=z;return;
}
inline int read(){
    char c;while(c=getchar(),c==' '||c=='\n');int data=c-48;
    while(c=getchar(),c>='0'&&c<='9')data=(data<<1)+(data<<3)+c-48;return data;
}
inline void DJ(int x){
    FOR(i,1,n)dis[i]=inf,vis[i]=0;
    dis[x]=0;
    heap.push(make_pair(0,x));
    int now;
    while(!heap.empty()){
        now=heap.top().second;
        heap.pop();
        if(vis[now])continue;
        vis[now]=1;
        VIS(now)
            if(dis[to[e]]>dis[now]+w[e]){
                dis[to[e]]=dis[now]+w[e];
                heap.push(make_pair(-dis[to[e]],to[e]));
            }
    }
}
inline void dfs(int now,int sum){
    if(sum+dis[now]>dis[1]+k)
        return;
    if(now==n){
        if(sum<=dis[1]+k)
            ++ans;
        ans%=p;
        return;
    }
    VIS(now){
        dfs(to[e],sum+w[e]);
    }
}
inline void solve0(){
    FOR(i,1,m){
        x[i]=read();
        y[i]=read();
        z[i]=read();
        add(x[i],y[i],z[i]);
    }
    DJ(1);
    int now;
    while(!heap.empty())heap.pop();
    FOR(i,1,n)vis[i]=0,dp[i]=0ll;  
    dp[1]=1ll;
    heap.push(make_pair(0,1));
    while(!heap.empty()){
        now=heap.top().second;
        heap.pop();
        if(vis[now])continue;
        vis[now]=1;
        VIS(now)
            if(dis[to[e]]==dis[now]+w[e]){
                dp[to[e]]+=dp[now];
                dp[to[e]]%=p;
                heap.push(make_pair(-dis[to[e]],to[e]));
            }
    }
    printf("%d\n",dp[n]);
}
int main(){
    T=read();
    die:;
    while(T--){
        n=read();m=read();k=read();p=read();
        tot=0;
        FOR(i,1,n)las[i]=0;
        if(k==0){
            solve0();
            continue;
        }
        FOR(i,1,m){
            x[i]=read();
            y[i]=read();
            z[i]=read();
            add(y[i],x[i],z[i]);
        }
        DJ(n);
        tot=0;
        FOR(i,1,n)las[i]=0,vis[i]=0;
        FOR(i,1,m)
            if(z[i]==0)
                add(x[i],y[i],z[i]);
        tot=0;
        FOR(i,1,n)las[i]=0,vis[i]=0;
        FOR(i,1,m)add(x[i],y[i],z[i]);
        ans=0;
        dfs(1,0);
        printf("%d\n",ans);
    }
    return 0;
}

　　

Day1 崩了 100+50+30=180

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