BestCoder Round #47 ($) (hdu 5280 , hdu 5281 , hdu 5282 , hdu 5283)

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1001 Senior’s Array

这题暴力枚举替换哪一个，然后dp就行。。O(n^2)的复杂度
这题一开始看错了，一直用线段树WA了好多发，好伤。

//author: CHC//First Edit Time:  2015-07-11 20:00#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <queue>#include <set>#include <algorithm>#include <limits>using namespace std;typedef long long LL;const int MAXN=1e+4;const int MAXM=1e+5;const int INF = numeric_limits<int>::max();const LL LL_INF= numeric_limits<LL>::max();LL A[MAXN]; LL B[MAXN]; LL dp[MAXN];int n,p,t;int main(){    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&p);        for(int i=1;i<=n;i++){            scanf("%I64d",&A[i]);            B[i]=A[i];        }        LL ans=p;        for(int i=1;i<=n;i++){            B[i]=(LL)p;            dp[0]=0;            for(int j=1;j<=n;j++){                dp[j]=max(B[j],dp[j-1]+B[j]);                ans=max(ans,dp[j]);            }            B[i]=A[i];        }        printf("%I64d\n",ans);    }    return 0;}

1002 Senior’s Gun

这题排序，然后攻击力最大的打防御力最小的，一个O(n)就可以了。
这题我数组范围开小了导致最后Judge大数据的时候WA了。

//author: CHC//First Edit Time:  2015-07-11 20:17#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <queue>#include <set>#include <algorithm>#include <limits>using namespace std;typedef long long LL;const int MAXN=100000+10;const int MAXM=1e+5;const int INF = numeric_limits<int>::max();const LL LL_INF= numeric_limits<LL>::max();LL A[MAXN],B[MAXN];int n,m;int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)scanf("%I64d",&A[i]);        for(int i=0;i<m;i++)scanf("%I64d",&B[i]);        sort(A,A+n);        sort(B,B+m);        LL sum=0;        for(int i=n-1,j=0;i>=0&&j<m;i--,j++){            if(A[i]>=B[j])sum+=A[i]-B[j];            else break;        }        printf("%I64d\n",sum);    }    return 0;}

1003 Senior’s String

这题当时我没有做出，其实题解写的很清楚了。

//author: CHC//First Edit Time:  2015-07-14 00:06#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <queue>#include <set>#include <algorithm>#include <limits>using namespace std;typedef long long LL;const int MAXN=1010;const int INF = numeric_limits<int>::max();const LL LL_INF= numeric_limits<LL>::max();const int mod=1e+9 + 7;char str1[MAXN],str2[MAXN];int dp[MAXN][MAXN],P[MAXN][MAXN],dp2[MAXN][MAXN];int main(){    int t;    scanf("%d",&t);    while(t--){        scanf(" %s %s",str1+1,str2+1);        int lenx=strlen(str1+1);        int leny=strlen(str2+1);        memset(dp,0,sizeof(dp));        for(int i=1;i<=lenx;i++)        for(int j=1;j<=leny;j++){            if(str1[i]==str2[j])dp[i][j]=dp[i-1][j-1]+1;            else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);        }        memset(P,0,sizeof(P));        for(int i=1;i<=lenx;i++)        for(int j=1;j<=leny;j++)            if(str1[i]==str2[j])P[i][j]=j;            else P[i][j]=P[i][j-1];        memset(dp2,0,sizeof(dp2));        for(int i=1;i<=lenx;i++)            for(int j=1;j<=leny;j++){                if(dp[i][j]==0){                    dp2[i][j]=1;                    continue;                }                if(dp[i][j]==dp[i-1][j])dp2[i][j]+=dp2[i-1][j];                dp2[i][j]%=mod;                if(P[i][j]&&dp[i][j]==dp[i-1][P[i][j]-1]+1){                    if(dp2[i-1][P[i][j]-1]==0)                        dp2[i][j]+=dp2[i-1][P[i][j]-1]+1;                    else dp2[i][j]+=dp2[i-1][P[i][j]-1];                }                dp2[i][j]%=mod;                //printf("%d %d %d %d\n",i,j,dp2[i][j],dp[i][j]==dp[i-1][j]);            }        printf("%d\n",dp2[lenx][leny]);    }    return 0;}

1004 Senior’s Fish

这题当时我也没做出来，后来看题解的时候第一感觉就是复杂度很高。后来才反应过来。。。过了的时候那种极度不科学的感觉那种冲击。。

//author: CHC//First Edit Time:  2015-07-14 13:41#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <queue>#include <set>#include <algorithm>#include <limits>using namespace std;typedef long long LL;const int MAXN=100000+100;const LL LL_INF=numeric_limits<LL>::max();#define lson L,mid,rt<<1#define rson mid+1,R,rt<<1|1long long X[4],Y[4];long long tx[MAXN],ty[MAXN];struct Tree {    long long maX[4],maY[4],addx,addy,sum[4];}tr[MAXN<<2];void pushup(int rt){    for(int i=0;i<4;i++){        tr[rt].maX[i]=max(tr[rt<<1].maX[i],tr[rt<<1|1].maX[i]);        tr[rt].maY[i]=max(tr[rt<<1].maY[i],tr[rt<<1|1].maY[i]);        tr[rt].sum[i]=tr[rt<<1].sum[i]+tr[rt<<1|1].sum[i];    }}void pushdown(int rt){    if(tr[rt].addx){        tr[rt<<1].addx+=tr[rt].addx;        tr[rt<<1|1].addx+=tr[rt].addx;        for(int i=0;i<4;i++){            tr[rt<<1].maX[i]+=tr[rt].addx;            tr[rt<<1|1].maX[i]+=tr[rt].addx;        }        tr[rt].addx=0;    }    if(tr[rt].addy){        tr[rt<<1].addy+=tr[rt].addy;        tr[rt<<1|1].addy+=tr[rt].addy;        for(int i=0;i<4;i++){            tr[rt<<1].maY[i]+=tr[rt].addy;            tr[rt<<1|1].maY[i]+=tr[rt].addy;        }        tr[rt].addy=0;    }}void build(int L,int R,int rt){    tr[rt].addy=tr[rt].addx=0;    if(L==R){        for(int i=0;i<4;i++){            tr[rt].maX[i]=tx[L];            tr[rt].maY[i]=ty[L];            tr[rt].sum[i]=1;        }        return ;    }    int mid=(L+R)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int L,int R,int rt,int l,int r,long long v,int flag){    if(l<=L&&R<=r){        if(flag==0){            for(int i=0;i<4;i++) tr[rt].maX[i]+=v;            tr[rt].addx+=v;        }        else {            for(int i=0;i<4;i++) tr[rt].maY[i]+=v;            tr[rt].addy+=v;        }        return ;    }    pushdown(rt);    int mid=(L+R)>>1;    if(l<=mid)update(lson,l,r,v,flag);    if(r>mid)update(rson,l,r,v,flag);    pushup(rt);}void adjust(int L,int R,int rt){    int mid=(L+R)>>1;    for(int i=0;i<4;i++){        if(tr[rt].maX[i]>X[i]||tr[rt].maY[i]>Y[i]){            if(L==R){                tr[rt].maX[i]=LL_INF+1;                tr[rt].maY[i]=LL_INF+1;                tr[rt].sum[i]=0;                continue;            }            pushdown(rt);            adjust(lson);adjust(rson);            pushup(rt);        }    }}LL query(int L,int R,int rt,int l,int r,int f){    if(l<=L&&R<=r)return tr[rt].sum[f];    int mid=(L+R)>>1;    LL ans=0;    if(l<=mid)ans+=query(lson,l,r,f);    if(r>mid)ans+=query(rson,l,r,f);    return ans;}int main(){    int n,m,t;    int xx1,xx2,yy1,yy2;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        scanf("%d%d%d%d",&xx1,&yy1,&xx2,&yy2);        X[0]=xx2;Y[0]=yy2;        X[1]=xx1-1;Y[1]=yy2;        X[2]=xx2;Y[2]=yy1-1;        X[3]=xx1-1;Y[3]=yy1-1;        for(int i=1;i<=n;i++){            scanf("%I64d%I64d",&tx[i],&ty[i]);        }        build(1,n,1);        int type,d,tl,tr;        scanf("%d",&m);        while(m--){            scanf("%d",&type);            if(type==1){                scanf("%d%d%d",&tl,&tr,&d);                update(1,n,1,tl,tr,(LL)d,0);            }            if(type==2){                scanf("%d%d%d",&tl,&tr,&d);                update(1,n,1,tl,tr,(LL)d,1);            }            if(type==3){                adjust(1,n,1);                scanf("%d%d",&tl,&tr);                LL ans=query(1,n,1,tl,tr,0)-query(1,n,1,tl,tr,1)-query(1,n,1,tl,tr,2)+query(1,n,1,tl,tr,3);                printf("%I64d\n",ans);            }        }    }    return 0;}

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