poj_2956

题目链接：http://poj.org/problem?id=2965

题意就是给你一个4*4的举证，包含'+'和'-'，可以任意改变其中一种符号，但改变时会把该符号所在的行和列全部都改变，问最少改变几次才能把所有的都变为 '-。

这题和poj1753很类似，故可以用枚举+dfs的方法来做。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,c;#define N  100100char ch[20][20];int mp[20][20];int vis[20][20];int step;int flag;vector<pair<int,int> > v;void Init(){for (int i = 0; i < 4; i++)for (int j = 0; j < 4; j++) {if (ch[i][j] == '-')mp[i][j] = 0;elsemp[i][j] = 1;}}bool check(){for(int i=0;i<4;i++)for(int j=0;j<4;j++)if(mp[i][j]!=0) return false;return true;}void flip(int row,int col){for(int i=0;i<4;i++)mp[row][i] = 1-mp[row][i];for(int i=0;i<4;i++){if(i==row) continue;mp[i][col] = 1-mp[i][col];}}void dfs(int row,int col,int dep){if(dep == step){flag = check();return;}if(flag) return;if(row == 4) return;if(col<3)dfs(row,col+1,dep);elsedfs(row+1,0,dep);if(flag) return;//!这个一定得写flip(row,col);//changevis[row][col] = !vis[row][col];if(col<3)dfs(row,col+1,dep+1);elsedfs(row+1,0,dep+1);if(flag) return;//!这个一定得写flip(row,col);vis[row][col] = !vis[row][col];return;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);#endif    while(sfs(ch[0])!=EOF){    for(int i=1;i<4;i++)    sfs(ch[i]);    Init();    flag=0;    memset(vis,0,sizeof vis);    for(step=0;step<=16;step++){    dfs(0,0,0);    if(flag)     break;    }    if(flag){    printf("%d\n",step);    for(int i=0;i<4;i++)    for(int j=0;j<4;j++){    if(vis[i][j]){    printf("%d %d\n",i+1,j+1);    }    }    }    }return 0;}

对于这道题，我们还有一种做法，就是对于一个'+'号，我们怎么只把它变为‘-’,那就是把它所在的行和列中所有的元素都变换一次，这样子，如果到最后某个位置的符号被变换了奇数次，那么它就是属于被改变的位置。最后统计输出即可。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,c;#define N  110char ch[20][20];int vis[20][20];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);#endif    while(sfs(ch[0])!=EOF){    for(int i=1;i<4;i++)    sfs(ch[i]);    memset(vis,0,sizeof vis);    for(int i=0;i<4;i++){    for(int j=0;j<4;j++){    if(ch[i][j] == '+'){    for(int s=0;s<4;s++)    vis[i][s]++;    for(int s=0;s<4;s++)    vis[s][j]++;    vis[i][j]--;<span style="white-space:pre"></span>//i,j处被改变了两次，所以得减一次    }    }    }    int ans=0;    for(int i=0;i<4;i++)    for(int j=0;j<4;j++)<span style="white-space:pre"></span>    ans += (vis[i][j]%2);<span style="white-space:pre"></span>//统计被改变奇数次的位置    printf("%d\n",ans);    for(int i=0;i<4;i++)    for(int j=0;j<4;j++){    if(vis[i][j]%2)    printf("%d %d\n",i+1,j+1);    }    }return 0;}