UVA-1030-Image Is Everything

来源：互联网发布：软件开发精品课程编辑：程序博客网时间：2024/06/06 01:28
这个题先找出“看穿”的所有方块，拿下来，再不断找每一块在各个视图中颜色不一样的给删掉，直到没有为止，本题难点在于各个视图和小方块的坐标转化，还是看看刘汝佳的代码，太屌了，我用了两种转化，他一种就搞出来了，以下是我的代码：
#include<stdio.h>#include<iostream>using namespace std;void kanchuan(char tu[][15][15],int m,int ti[][15][15],int n){    for(int i = 0;i < n;i++)        for(int j = 0;j < n;j++)        {            if(tu[m][i][j] == '.' && m == 0) for(int k = 0;k < n;k++) ti[j][k][n - i - 1] = 0;            if(tu[m][i][j] == '.' && m == 1) for(int k = 0;k < n;k++) ti[k][j][n - i - 1] = 0;            if(tu[m][i][j] == '.' && m == 2) for(int k = 0;k < n;k++) ti[n - j - 1][k][n - i - 1] = 0;            if(tu[m][i][j] == '.' && m == 3) for(int k = 0;k < n;k++) ti[k][n - j - 1][n - i - 1] = 0;            if(tu[m][i][j] == '.' && m == 4) for(int k = 0;k < n;k++) ti[j][i][k] = 0;            if(tu[m][i][j] == '.' && m == 5) for(int k = 0;k < n;k++) ti[j][n - i - 1][k] = 0;        }}int main(){    int n;    while(cin>>n && n)    {        char tu[6][15][15];        int ti[15][15][15],ans = 0;        for(int i = 0; i < n; i++)        {            scanf("%s",tu[0][i]);            scanf("%s",tu[1][i]);            scanf("%s",tu[2][i]);            scanf("%s",tu[3][i]);            scanf("%s",tu[4][i]);            scanf("%s",tu[5][i]);        }        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                for(int k = 0; k < n; k++) ti[i][j][k] = 1;        kanchuan(tu,0,ti,n); //前视图        kanchuan(tu,1,ti,n); //左视图        kanchuan(tu,2,ti,n); //后视图        kanchuan(tu,3,ti,n); //右视图        kanchuan(tu,4,ti,n); //顶视图        kanchuan(tu,5,ti,n); //底视图        while(1)        {            int flag = 0;  //0则正确            for(int i = 0; i < n; i++) //检测某个面是否应该被删除                for(int j = 0; j < n; j++)                    for(int k = 0; k < n; k++)                    if(ti[i][j][k])                    {                        char color = '0';                        //检测可以对比扫描的面                        int up = 1,down = 1,left = 1,right = 1,near = 1,notnear = 1;  //1是未被挡住                        for(int l = 0; l < i; l++) if(ti[l][j][k]) left = 0;                        for(int l = i + 1; l < n; l++) if(ti[l][j][k]) right = 0;                        for(int l = 0; l < j; l++) if(ti[i][l][k]) notnear = 0;                        for(int l = j + 1; l < n; l++) if(ti[i][l][k]) near = 0;                        for(int l = 0; l < k; l++) if(ti[i][j][l]) down = 0;                        for(int l = k + 1; l < n; l++) if(ti[i][j][l]) up = 0;                        //在各个视图中找到这个小方块，并对比                        if(up && color == '0')   color = tu[4][j][i];                        if(down && color == '0') color = tu[5][n - j - 1][i];                        if(left && color == '0') color = tu[1][n - k - 1][j];                        if(right && color == '0') color = tu[3][n - k - 1][n  - j - 1];                        if(near && color == '0') color = tu[0][n - k - 1][i];                        if(notnear && color == '0') color = tu[2][n - k - 1][n - i - 1];                        if(up && color !=tu[4][j][i])                        {                            flag = 1;                            ti[i][j][k] = 0;                            continue;                        }                        if(down && color !=tu[5][n - j - 1][i])                        {                            flag = 1;                            ti[i][j][k] = 0;                            continue;                        }                        if(left && color != tu[1][n - k - 1][j])                        {                            flag = 1;                            ti[i][j][k] = 0;                            continue;                        }                        if(right && color !=tu[3][n - k - 1][n - j - 1])                        {                            flag = 1;                            ti[i][j][k] = 0;                            continue;                        }                        if(near && color !=tu[0][n - k - 1][i])                        {                            flag = 1;                            ti[i][j][k] = 0;                            continue;                        }                        if(notnear && color !=tu[2][n - k - 1][n - i - 1])                        {                            flag = 1;                            ti[i][j][k] = 0;                            continue;                        }                    }            if(!flag) break;        }        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                for(int k = 0; k < n; k++) if(ti[i][j][k]) ans++;        printf("Maximum weight: %d gram(s)\n",ans);    }    return 0;}
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