[NOIP2011TG] Day 1 Solution

T1:由n至1判断是否cover到即可。

Code:

#include<cstdio> 
#define MN 10005 
using namespace std; 
int n,x,y,a[MN],b[MN],g[MN],k[MN]; 
int main() 
{ 
    scanf("%d",&n); 
    for (int i=1;i<=n;i++) 
    scanf("%d%d%d%d",&a[i],&b[i],&g[i],&k[i]); 
    scanf("%d%d",&x,&y); 
    for (int i=n;i>=1;i--){ 
        if (x>=a[i]&&x<=a[i]+g[i]&&y>=b[i]&&y<=b[i]+k[i]) 
        {printf("%d",i);return 0;} 
    } printf("-1");return 0; 
}

T2:

t[c]表示颜色为c的酒店有几间，cnt[c]表示当前可去的颜色为c的酒店有几间。

读入每间酒店的色调和咖啡店的最低消费同时，若l<=p，则以前相同色调的酒店都可以住，更新cnt,答案加上cnt[c].

Code:

#include<iostream>  
#include<cmath>  
#include<cstdio>  
#include<cstdlib>  
#include<cstring>  
#include<queue>  
#include<algorithm> 
using namespace std; 
int n,k,p,c,l,ans; 
int cnt[52],t[52]; 
int main() 
{ 
    scanf("%d%d%d",&n,&k,&p); 
    for (int i=1;i<=n;i++){ 
        scanf("%d%d",&c,&l); 
        if (l<=p)memcpy(cnt,t,sizeof(t)); 
        ans+=cnt[c]; 
        if (l<=p) cnt[c]++;t[c]++; 
    } 
    printf("%d",ans);return 0; 
}

T3:

dfs枚举交换的位置，并模拟消除与掉落的过程（可能有多次），将空白当作颜色0处理。

剪枝1：若某颜色方块少于三个，return;

剪枝2：不必分1或-1的情况分别讨论，因为若两个不为0的色块位置互换，可看成是左面色块右移1得到的（1的优先级大于-1）。若左边色块为0，则为-1,右边的色块为0，则为1.

Code:

#include<cstdio> 
#include<cstring> 
#include<algorithm> 
using namespace std; 
int x[6],y[6],dir[6],ct[11]; 
int map[6][8],mcol,n,t,cnt=0; 
bool check() 
{ 
    for (int i=1;i<=5;i++) 
    for (int j=1;j<=7;j++) if (map[i][j]) return 0; 
    return 1; 
} 
void print() 
{ 
    for (int i=1;i<=n;i++) printf("%d %d %d\n",x[i]-1,y[i]-1,dir[i]);exit(0); 
} 
void down() 
{ 
    for (int i=1;i<=5;i++){ 
        int pos=0; 
        for (int j=1;j<=7;j++) 
        if (map[i][j]){ 
            int temp=map[i][j]; 
            map[i][j]=0;map[i][++pos]=temp;  
        } 
    } 
} 
bool cancel() 
{ 
    bool used[6][8]={0},changed=0; 
    for (int i=1;i<=5;i++) 
    for (int j=1;j<=7;j++){ 
        if (!map[i][j]) continue; 
        if (i<=3&&map[i][j]==map[i+1][j]&&map[i][j]==map[i+2][j]) 
        {used[i][j]=used[i+1][j]=used[i+2][j]=1;changed=1;} 
        if (j<=5&&map[i][j]==map[i][j+1]&&map[i][j]==map[i][j+2]) 
        {used[i][j]=used[i][j+1]=used[i][j+2]=1;changed=1;} 
    } 
    for (int i=1;i<=5;i++) 
    for (int j=1;j<=7;j++) if (used[i][j]) map[i][j]=0;   
    return changed; 
} 
void dfs(int step) 
{ 
    int tmap[6][8]; 
    if (step>n) {if (check()) print();else return;} 
    for (int i=1;i<=5;i++) 
    for (int j=1;j<=7;j++) ct[map[i][j]]++; 
    for (int i=1;i<=mcol;i++) if(ct[i]==1||ct[i]==2) return; 
    memcpy(tmap,map,sizeof(map)); 
    for (int i=1;i<5;i++) 
    for (int j=1;j<=7;j++) 
    { 
        if (map[i][j]!=map[i+1][j]){ 
            if (!map[i][j]) { 
                x[step]=i+1;y[step]=j;dir[step]=-1; 
            }else{ 
                x[step]=i;y[step]=j;dir[step]=1; 
            } 
            swap(map[i][j],map[i+1][j]); 
            down();while(cancel())down();dfs(step+1); 
            memcpy(map,tmap,sizeof(map)); 
        } 
    } 
} 
int main() 
{ 
    scanf("%d",&n);memset(map,0,sizeof(map)); 
    for (int i=1;i<=5;i++) 
    { 
        int tmp=0;scanf("%d",&t);mcol=t; 
        while(t){map[i][++tmp]=t;scanf("%d",&t);mcol=max(t,mcol);} 
    } 
    dfs(1);printf("-1"); 
    return 0; 
}