HDU 5755 Gambler Bo

来源：互联网发布：什么是软件开发模式编辑：程序博客网时间：2024/05/17 22:50

解题思路：

这题可以对每个位置设个未知数，然后构造方程组高斯消元求解，但是这样复杂度O(（n*m）^3)

代码一：

#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>//#pragma comment(linker, "/STACK:1024000000,1024000000");using namespace std;#define INF 0x3f3f3f3fconst int MOD = 3;const int MAXN = 905;int a[MAXN][MAXN];int x[MAXN];int Gauss(int equ, int var) {    int row, max_r, col;    for(row = 0, col = 0; col < var && row < equ; col++, row++) {        max_r = row;        for(int i = row + 1; i < equ; i++) {            if(abs(a[i][col]) > abs(a[max_r][col])) {                max_r = i;            }        }        if(!a[max_r][col]) {            row--;            continue;        }        if(row != max_r) {            for(int j = 0; j <= var; j++) {                swap(a[row][j], a[max_r][j]);            }        }        for(int i = row + 1; i < equ; i++) {            if(a[i][col]) {                int lcm = a[i][col] / __gcd(a[i][col], a[row][col]) * a[row][col];                int t1 = lcm / a[i][col];                int t2 = lcm / a[row][col];                for(int j = col; j <= var; j++) {                    a[i][j] = ((a[i][j] * t1 - a[row][j] * t2) % MOD + MOD) % MOD;                }            }        }    }    for(int i = var - 1; i >= 0; i--) {        x[i] = a[i][var];        for(int j = i + 1; j < var; j++) {            x[i] = ((x[i] - a[i][j] * x[j]) % MOD + MOD) % MOD;        }        x[i] = a[i][i] * x[i] % MOD;    }}int n,m;int mp[50][50];int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)                scanf("%d",&mp[i][j]);            memset(a,0,sizeof a);            for(int j=0;j<n*m;j++)            {                int u=j-m,d=j+m,l=j-1,r=j+1;                a[j][j]=2;                if(u>=0) a[j][u]=1;                if(d<n*m) a[j][d]=1;                if(j%m>0) a[j][l]=1;                if(j%m<m-1) a[j][r]=1;                a[j][n*m]=2*mp[j/m][j%m]%MOD;            }        int ans=Gauss(n*m,n*m);        vector<pair<int,int> > vec;        for(int i=0;i<n*m;i++)        {            for(int j=0;j<x[i];j++)                vec.push_back(pair<int,int>(i/m+1,i%m+1));        }        printf("%d\n",vec.size());        for(int i=0;i<vec.size();i++)        {            printf("%d %d\n",vec[i].first,vec[i].second);        }    }    return 0;}

思路二：

其实这题可以只设出第一行的操作次数，然后后面的行的操作必须保证前一行复位，这样用dp算出最后最后一行的系数，dp[i][j][k]表示第i行中第j个数对第k个变量的系数，就可以根据最后一行复位来列方程，进而高斯消元求解，复杂度为O(m^3)。

代码二：

#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>//#pragma comment(linker, "/STACK:1024000000,1024000000");using namespace std;#define INF 0x3f3f3f3fconst int MOD = 3;const int MAXN = 35;int a[MAXN][MAXN];int x[MAXN];int Gauss(int equ, int var) {    int row, max_r, col;    for(row = 0, col = 0; col < var && row < equ; col++, row++) {        max_r = row;        for(int i = row + 1; i < equ; i++) {            if(abs(a[i][col]) > abs(a[max_r][col])) {                max_r = i;            }        }        if(!a[max_r][col]) {            row--;            continue;        }        if(row != max_r) {            for(int j = 0; j <= var; j++) {                swap(a[row][j], a[max_r][j]);            }        }        for(int i = row + 1; i < equ; i++) {            if(a[i][col]) {                int lcm = a[i][col] / __gcd(a[i][col], a[row][col]) * a[row][col];                int t1 = lcm / a[i][col];                int t2 = lcm / a[row][col];                for(int j = col; j <= var; j++) {                    a[i][j] = ((a[i][j] * t1 - a[row][j] * t2) % MOD + MOD) % MOD;                }            }        }    }    for(int i = var - 1; i >= 0; i--) {        x[i] = a[i][var];        for(int j = i + 1; j < var; j++) {            x[i] = ((x[i] - a[i][j] * x[j]) % MOD + MOD) % MOD;        }        x[i] = a[i][i] * x[i] % MOD;    }}int n,m;int mp[50][50];int dp[50][50][50];int dp2[50][50];int ansx[1805],ansy[1805];int vis[35];int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof dp);        memset(dp2,0,sizeof dp2);        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                scanf("%d",&mp[i][j]),dp2[i][j]=mp[i][j];        for(int i=1; i<=m; i++)        {            dp[1][i][i]=2;            dp[1][i-1][i]=1;            dp[1][i+1][i]=1;            dp[2][i][i]+=1;        }        for(int i=2; i<=n; i++)            for(int j=1; j<=m; j++)                for(int k=1; k<=m; k++)                {                    int cc=dp[i-1][j][k];                    dp[i][j][k]+=cc;dp[i][j][k]%=MOD;                    dp[i][j-1][k]+=2*cc;dp[i][j-1][k]%=MOD;                    dp[i][j+1][k]+=2*cc;dp[i][j+1][k]%=MOD;                    dp[i+1][j][k]+=2*cc;dp[i+1][j][k]%=MOD;                }        for(int i=2; i<=n; i++)        {            for(int j=1; j<=m; j++)            {                int cc=dp2[i-1][j];                dp2[i][j]+=cc;dp2[i][j]%=MOD;                dp2[i][j-1]+=2*cc;dp2[i][j-1]%=MOD;                dp2[i][j+1]+=2*cc;dp2[i][j+1]%=MOD;                dp2[i+1][j]+=2*cc;dp2[i+1][j]%=MOD;            }        }        for(int i=0; i<m; i++)            for(int j=0; j<m; j++)            {                a[i][j]=dp[n][i+1][j+1]%MOD;                a[i][j]=(a[i][j]+MOD)%MOD;            }        for(int i=0; i<m; i++)        {            a[i][m]=2*(dp2[n][i+1])%MOD;            a[i][m]=(a[i][m]+MOD)%MOD;        }        int ans=Gauss(m,m);        int cnt=0;        for(int i=0; i<m; i++)        {            x[i]=(x[i]%MOD+MOD)%MOD;            mp[1][i+1]+=2*x[i]; mp[1][i+1]%=MOD;            mp[1][i]+=x[i]; mp[1][i]%=MOD;            mp[1][i+2]+=x[i];mp[1][i+2]%=MOD;            mp[2][i+1]+=x[i]; mp[2][i+1]%=MOD;            for(int j=1;j<=(x[i]%MOD+MOD)%MOD;j++)            {                ansx[cnt]=1;                ansy[cnt++]=i+1;            }        }        for(int i=2; i<=n; i++)        {            for(int j=1; j<=m; j++)            {                mp[i-1][j]%=3;                if(mp[i-1][j]==1)                {                    mp[i-1][j]+=2;mp[i-1][j]%=MOD;                    mp[i][j]=(mp[i][j]+4)%3;                    mp[i][j-1]=(mp[i][j-1]+2)%3;                    mp[i][j+1]+=2;mp[i][j+1]%=MOD;                    mp[i+1][j]+=2;mp[i+1][j]%=MOD;                    ansx[cnt]=i;                    ansy[cnt++]=j;                    ansx[cnt]=i;                    ansy[cnt++]=j;                }                else if(mp[i-1][j]==2)                {                    mp[i-1][j]+=1;mp[i-1][j]%=MOD;                    mp[i][j]=(mp[i][j]+2)%3;                    mp[i][j-1]=(mp[i][j-1]+1)%3;                    mp[i][j+1]+=1;mp[i][j+1]%=MOD;                    mp[i+1][j]+=1;mp[i+1][j]%=MOD;                    ansx[cnt]=i;                    ansy[cnt++]=j;                }            }        }        printf("%d\n",cnt);        for(int i=0;i<cnt;i++)            printf("%d %d\n",ansx[i],ansy[i]);    }    return 0;}

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