【NOIP模拟8.8】

T1 expedition

离散化后记录每一种士兵的出现次数计算即可

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define LL long longusing namespace std;LL n,a[100010],b[100010],c[100010];LL cnt[100010],ans;int main(){    freopen("expedition.in","r",stdin);    freopen("expedition.out","w",stdout);    scanf("%lld",&n);    for(int i=1;i<=n;i++)   scanf("%lld%lld",&a[i],&b[i]),c[i]=b[i];    sort(c+1,c+n+1);    int m=unique(c+1,c+n+1)-c-1;    for(int i=1;i<=n;i++)   b[i]=lower_bound(c+1,c+m+1,b[i])-c,cnt[b[i]]+=a[i];    for(int i=1;i<=n;i++)   ans+=cnt[b[i]]*cnt[b[i]],cnt[b[i]]=0;    printf("%lld",ans);    return 0;}

T2 simplify

相当于一个多项式的中缀表达式求值，一个数组其实就可以用来表示一个多项式，数组的每一项是多项式每一项的系数，所以只需要手动模拟一下多项式加法，减法，乘法，递归计算的复杂度O(n4),如果用两个栈分别记录运算的多项式和符号，根据符号的优先级计算，复杂度是O(n3)
PS：拉格朗日插值法可以做到O(n2)，但实际上和上述算法时间是接近的

//by sdfzchy#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=1010,mod=10007;char s[1010];int len,optop,top,op[1010];struct info{    int len,a[501];    info(){len=0,memset(a,0,sizeof(a));}    void reset(){while(len&&!a[len]) len--;}}stack[1010];info operator + (info a,info b){    info c;    c.len=max(a.len,b.len);    for(int i=0;i<=c.len;i++) c.a[i]=a.a[i]+b.a[i];    c.reset();    return c;   }info operator - (info a,info b){    info c;    c.len=max(a.len,b.len);    for(int i=0;i<=c.len;i++) c.a[i]=a.a[i]-b.a[i];    c.reset();    return c;}info operator * (info a,info b){    info c;    c.len=a.len+b.len;    for(int i=0;i<=a.len;i++)        for(int j=0;j<=b.len;j++)            (c.a[i+j]+=a.a[i]*b.a[j])%=mod;    c.reset();    return c;       }int prio(char a){    if(a=='+'||a=='-') return 1;    if(a=='*') return 2;    if(a=='(') return 0;}void calc(){    if(op[optop]=='+') stack[top-1]=stack[top-1]+stack[top];    else if(op[optop]=='-') stack[top-1]=stack[top-1]-stack[top];    else if(op[optop]=='*') stack[top-1]=stack[top-1]*stack[top];    optop--;top--;}void pause(){return;}int main(){    freopen("simplify.in","r",stdin);    freopen("simplify.out","w",stdout);    scanf("%s",s+1);    len=strlen(s+1);    for(int i=1;i<=len;i++)    {        if(s[i]=='(')   op[++optop]='(';        else if(s[i]==')')        {            pause();            while(op[optop]!='(') calc();            optop--;        }        else if(s[i]=='x')        {            info a;            a.a[1]=1;            a.len=1;            memset(stack[top+1].a,0,sizeof(stack[top+1].a));            stack[++top]=a;        }        else if(s[i]>='0'&&s[i]<='9')        {            info a;            int x=0;            while(s[i]>='0'&&s[i]<='9') x=x*10+s[i++]-'0';            i--;            a.a[0]=x;            memset(stack[top+1].a,0,sizeof(stack[top+1].a));            stack[++top]=a;        }        else if(s[i]=='+'||s[i]=='-')        {            if(i==1||(s[i-1]=='('))            {                int flag=(s[i++]=='+')?1:(-1);                if(s[i]=='x')                {                    info a;                    a.a[1]=flag;                    a.len=1;                    memset(stack[top+1].a,0,sizeof(stack[top+1].a));                        stack[++top]=a;                }                if(s[i]>='0'&&s[i]<='9')                {                    info a;                    int x=0;                    while(s[i]>='0'&&s[i]<='9') x=x*10+s[i++]-'0';                    i--;                    a.a[0]=x*flag;                    memset(stack[top+1].a,0,sizeof(stack[top+1].a));                    stack[++top]=a;                }            }            else             {                while(prio(s[i])<=prio(op[optop])) calc();                op[++optop]=s[i];            }        }        else if(s[i]=='*')        {            while(prio(s[i])<=prio(op[optop])) calc();            op[++optop]=s[i];        }    }    while(optop) calc();    cout<<stack[1].len<<endl;    for(int i=0;i<=stack[1].len;i++)        cout<<(stack[1].a[i]%mod+mod)%mod<<endl;    return 0;}

T3 production

正反向各建一次图跑SPFA即可

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define LL long longusing namespace std;const int N=3010;const int M=100010;int head[N],ecnt,eecnt,hhead[N],n,m,dis[N],ddis[N],q;bool vis[N];struct edge{    int to,nxt,dis;}e[M],ee[M];void add_edge(int u,int v,int w){    e[++ecnt].nxt=head[u];    e[ecnt].to=v;    e[ecnt].dis=w;    head[u]=ecnt;    ee[++eecnt].nxt=hhead[v];    ee[eecnt].to=u;    ee[eecnt].dis=w;    hhead[v]=eecnt;}void SPFA(){    int v,u;    queue<int> q;    q.push(1);    dis[1]=0;    vis[1]=1;    while(!q.empty())    {        u=q.front();        q.pop();        vis[u]=0;        for(int i=head[u];i;i=e[i].nxt)        {            v=e[i].to;            if(dis[u]+e[i].dis<dis[v])            {                dis[v]=dis[u]+e[i].dis;                if(!vis[v])                {                    q.push(v);                    vis[v]=1;                }            }        }    }}void SSPFA(){    int v,u;    queue<int> q;    q.push(1);    ddis[1]=0;    vis[1]=1;    while(!q.empty())    {        u=q.front();        q.pop();        vis[u]=0;        for(int i=hhead[u];i;i=ee[i].nxt)        {            v=ee[i].to;            if(ddis[u]+ee[i].dis<ddis[v])            {                ddis[v]=ddis[u]+ee[i].dis;                if(!vis[v])                {                    q.push(v);                    vis[v]=1;                }            }        }    }}int main(){    freopen("production.in","r",stdin);    freopen("production.out","w",stdout);    scanf("%d%d",&n,&m);    for(int i=1,u,v,w;i<=m;i++)    {        scanf("%d%d%d",&u,&v,&w);        add_edge(u,v,w);    }    memset(dis,0x3f,sizeof(dis));    memset(ddis,0x3f,sizeof(ddis));    SPFA();    SSPFA();    scanf("%d",&q);//  for(int i=1;i<=n;i++) cout<<dis[i]<<" "<<ddis[i]<<endl;    for(int i=1,u,v;i<=q;i++)    {        scanf("%d%d",&u,&v);        int tmp=ddis[u]+dis[v];        if(tmp>=1061109567) printf("-1\n");        else printf("%d\n",tmp);    }    return 0;}

PS:一套水题…