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P1074 靶形数独 (搜索决策)

时间:2020-03-06 23:57:56      阅读:110      评论:0      收藏:0      [点我收藏+]

标签:dfs   math   pair   end   its   ack   ref   operator   stream   

题目链接:https://www.luogu.com.cn/problem/P1074

 

详细讲解:https://www.luogu.com.cn/blog/cpp/solution-p1074

当暴力过不了的时候也可以考虑改变搜索的起点从而减少搜索树的大小

 

#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <cstring>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back

const double eps = 1e-10;
const int maxn = 10 + 10;
const LL mod = 1e9 + 7;
const LL INF = 1e18;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

struct Point {
    int r,sum;
    bool operator < (const Point &t) const {
        return sum < t.sum;
    }
}p[maxn];

int row[maxn][10],col[maxn][10],G[maxn][10];
int cnt,ans;
int a[maxn][maxn];
int mapp[maxn][4];

inline int get_which(int i,int j) {
    if (i <= 3) {
        if (j <= 3) return 1;
        else if (j <= 6) return 2;
        else return 3;
    } else if (i <= 6) {
        if (j <= 3) return 4;
        else if (j <= 6) return 5;
        else return 6;
    } else {
        if (j <= 3) return 7;
        else if (j <= 6) return 8;
        else return 9;
    }
}

inline int get_point(int i,int j) {
    if(i==1||j==1||i==9||j==9)   return 6;
    if(i==2||j==2||i==8||j==8)     return 7;
    if(i==3||j==3||i==7||j==7)   return 8;
    if(i==4||j==4||i==6||j==6)   return 9;
    return 10;
}

inline void dfs(int x,int sc) {
    if (x == cnt+1) {
        if (sc > ans)
            ans = sc;
        return ;
    }
    for (int i = 1;i <= 9;i++) {
        if (!row[mapp[x][0]][i] && !col[mapp[x][1]][i] && !G[mapp[x][3]][i]) {
            row[mapp[x][0]][i] = col[mapp[x][1]][i] = G[mapp[x][3]][i] = 1;
            dfs(x+1,sc+mapp[x][2]*i);
            row[mapp[x][0]][i] = col[mapp[x][1]][i] = G[mapp[x][3]][i] = 0;
        }
    }
}

int main() {
    int now = 0;
    for (int i = 1;i <= 9;i++) {
        for (int j = 1;j <= 9;j++) {
            p[i].r = i;
            cin >> a[i][j];
            if (a[i][j] > 0) {
                row[i][a[i][j]] = col[j][a[i][j]] = G[get_which(i, j)][a[i][j]] = 1;
                now += a[i][j] * get_point(i,j);
            }
            else
                p[i].sum++;
        }
    }
    sort(p+1,p+1+9);
    cnt = 0;
    for (int i = 1;i <= 9;i++) {
        for (int j = 1;j <= 9;j++) {
            if (a[p[i].r][j] == 0) {
                ++cnt;
                mapp[cnt][0] = p[i].r;
                mapp[cnt][1] = j;
                mapp[cnt][2] = get_point(p[i].r,j);
                mapp[cnt][3] = get_which(p[i].r,j);
            }
        }
    }
    ans = -1;
    dfs(1,now);
    cout << ans << endl;
    return 0;
}

 

P1074 靶形数独 (搜索决策)

标签:dfs   math   pair   end   its   ack   ref   operator   stream   

原文地址:https://www.cnblogs.com/-Ackerman/p/12431500.html

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