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Bzoj 1085: [SCOI2005]骑士精神

题目链接:https://www.lydsy.com/JudgeOnline/problem.php?id=1085
dfs + 剪枝.
剪枝方法:
1.每次交换只能改变一个位置.若发现之间相差的步数加上以前走的步数大于15的话,直接舍弃这一状态.
2.初始时,\(ans\)设为\(16\)
有了上面两个剪枝就A了.
照这节奏,SCOI2005就刷完了???

#include <iostream>
#include <cstdio>
#define X 6
using namespace std;

const int gx[] = {0,-2,-2,-1,-1,1,1,2,2};
const int gy[] = {0,-1,1,-2,2,-2,2,-1,1};

char map[X][X];
char c[X][X] = {
    '0','0','0','0','0','0',
    '0','1','1','1','1','1',
    '0','0','1','1','1','1',
    '0','0','0','*','1','1', 
    '0','0','0','0','0','1',
    '0','0','0','0','0','0'
};

int ans;

int inint(){
    int num = 0;
    for(int i = 1;i <= 5;++ i){
        for(int j = 1;j <= 5;++ j){
            if(c[i][j] != map[i][j])num ++;
        }
    }
    return num;
}

void dfs(int x,int y,int d,int tmp){
    int l = inint();
    if(d + l > 16)return;
    if(d > ans)return;
    if(l == 0) ans = d;
    
    for(int i = 1;i <= 8;++ i){
        if(x + gx[i] < 1 || x + gx[i] > 5)continue;
        if(y + gy[i] < 1 || y + gy[i] > 5)continue;
        if(tmp + i == 9)continue;
        swap(map[x][y],map[x + gx[i]][y + gy[i]]);
        dfs(x + gx[i],y + gy[i],d + 1,i);
        swap(map[x][y],map[x + gx[i]][y + gy[i]]);
    }
}

void work(){
    int x,y;
    for(int i = 1;i <= 5;++ i)cin >> map[i] + 1;
    for(int i = 1;i <= 5;++ i){
        for(int j = 1;j <= 5;++ j){
            if(map[i][j] == '*')
                x = i,y = j;
        }
    }
    ans = 16;
    dfs(x,y,0,0);
    printf("%d\n",ans == 16 ? -1 : ans);
    return;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t --){
        work();
    }
    return 0;
}

Bzoj 1085: [SCOI2005]骑士精神 (dfs)

标签：ini   scanf   inline   bzoj   int   tmp   https   直接   剪枝   

原文地址：https://www.cnblogs.com/tpgzy/p/9726569.html