hiho 31 扫雷二

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问题

http://hihocoder.com/contest/hiho31/problem/1

解法

首先要理解题意，共有三种规则，这三种规则要独立使用。每个规则都是应用在输入数据上求得结果。
第三种规则两两相交有24种情况。

#include <bits/stdc++.h>using namespace std;enum{maxn = 200+5, isB = 9, notB = 10};int G[maxn][maxn];int res[maxn][maxn];int n, m;int Delt[8][2] = {{-1, -1}, {0, -1}, {1, -1},                    {-1, 0},  {1, 0},{-1, 1}, {0, 1}, {1, 1}};int d2[12][2] = {{2, -2},{2, -1},{2, 0},                {1, -2},{1, -1},{1, 0},                {0, -2},{0, -1},                {-1, -2},{-1, -1},                {-2, -2},{-2, -1},};void caseA(int i, int j){    for (int c=0; c<8; c++)    {        int ii = i+Delt[c][0];        int jj = j +Delt[c][1];        if (G[ii][jj] ==-1)        {            res[ii][jj] = notB;        }    }}void caseB(int i, int j){    int nkNum = 0;    int bNum = 0;    for(int c=0; c<8; c++)    {        int ii = i+Delt[c][0];        int jj = j +Delt[c][1];        if (G[ii][jj] ==-1)            nkNum++;    }    if (nkNum == G[i][j])    {           for (int c =0; c <8; c++)        {            int ii = i+Delt[c][0];            int jj = j +Delt[c][1];            if (G[ii][jj] ==-1)            {                res[ii][jj] = isB;            }        }    }}int getUkNum(int i, int j, int k, int s){    int uknum =0;    for (int c =0; c<8; ++c)    {        int ii = i+Delt[c][0];        int jj = j+Delt[c][1];        if ((abs(ii-k) >1 || abs(jj-s)>1) && G[ii][jj]==-1)            uknum++;    }    return uknum;}void setUkToB(int i, int j, int k, int s){    for (int c =0; c<8; ++c)    {        int ii = i+Delt[c][0];        int jj = j+Delt[c][1];        if ((abs(ii-k) >1 || abs(jj-s)>1) && G[ii][jj]==-1)        {            res[ii][jj] = isB;        }    }}void test(int i, int j, int k, int s){    int ukNum1 = getUkNum(i, j, k, s);    int ukNum2 = getUkNum(k, s, i, j);    if (ukNum1 ==0 && ukNum2&& ukNum2 == G[k][s] - G[i][j])    {        setUkToB(k, s, i, j);    }else if (ukNum2 ==0 && ukNum1 && ukNum1 == G[i][j] - G[k][s])    {        setUkToB(i,j, k, s);    }}void caseC(int i, int j){    for (int c =0; c<12; ++c)    {        int ii = i+d2[c][0];        int jj = j +d2[c][1];        if (1<=ii&& ii<=n && 1<=jj&& jj<=m && 0<=G[ii][jj] && G[ii][jj]<=8)        {            test(i, j, ii, jj);        }    }}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d %d", &n, &m);        memset(res, 0, sizeof(res));        for (int i=0; i<=n+1; ++i)            G[i][0] = G[i][m+1] = notB;        for (int i=0; i<=m+1; ++i)            G[0][i] = G[n+1][i] = notB;        for (int i=1; i<=n; ++i)            for (int j=1; j<=m; ++j)                scanf("%d", &G[i][j]);        for (int i=1; i<=n; ++i)            for (int j=1; j<=m; ++j)            {                if (0<=G[i][j] && G[i][j]<=8)                    caseC(i, j);                if (G[i][j]==0)                    caseA(i,j);                if (1<=G[i][j])                    caseB(i, j);            }        int isNum = 0, notNum =0;        for (int i=1; i<=n; ++i)            for (int j=1; j<=m; ++j)                if (res[i][j] == isB)                    isNum++;                else if (res[i][j] == notB)                    notNum++;        printf("%d %d\n", isNum, notNum);    }    return 0;}

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