HDU4740-模拟

题目：题目链接

 

题意：就是驴子和老虎在一个森林中跑，要是他们相遇了就输出相遇坐标，否则输出-1.跑动的原则是不跑出森林并且不跑自己跑过的路

 

分析：刚刚开始用一直搜索找出驴子和老虎各自走的路径，结果华丽丽的爆栈了，如下：

 

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;#define maxn 1001int QuickMod(int  a,int b,int n){    int r = 1;    while(b)    {        if(b&1)            r = (r*a)%n;        a = (a*a)%n;        b >>= 1;    }    return r;}int vis1[maxn][maxn];int vis2[maxn][maxn];int n;int st1x, st1y, dir1;int st2x, st2y, dir2;int fp1;int fp2;int dirx[4] = {0, 1, 0, -1};int diry[4] = {1, 0, -1, 0};bool inmap(int x, int y){    if(x < 0 || x >= n || y < 0 || y >= n)        return false;    return true;}int xx1, yy1;int flag;int xx2, yy2;struct node{    int x, y;} num[maxn], cop[maxn];int cntx;int cnty;void DFSX(int x1, int y1, int d1){    if(fp1)        return;    vis1[x1][y1] = 1;    int xp = x1 + dirx[d1];    int yp = y1 + diry[d1];    if(inmap(xp, yp) && !vis1[xp][yp])    {        xx1 = xp;        yy1 = yp;        cntx ++;        num[cntx].x = xx1;        num[cntx].y = yy1;        DFSX(xx1, yy1, d1);    }    else    {        if(d1 == 3)        {            int tx = x1 + dirx[0];            int ty = y1 + diry[0];            if(!vis1[tx][ty] && inmap(tx, ty))            {                xx1 = tx;                yy1 = ty;                cntx ++;                num[cntx].x = xx1;                num[cntx].y = yy1;                DFSX(xx1, yy1, 0);            }            else                fp1 = 1;        }        else        {            int tx = x1 + dirx[d1+1];            int ty = y1 + diry[d1+1];            if(!vis1[tx][ty] && inmap(tx, ty))            {                xx1 = tx;                yy1 = ty;                cntx ++;                num[cntx].x = xx1;                num[cntx].y = yy1;                DFSX(xx1, yy1, d1+1);            }            else                fp1 = 1;        }    }    if(fp1)        return;}void DFSY(int x1, int y1, int d1){    if(fp2)        return;    vis2[x1][y1] = 1;    int xp = x1 + dirx[d1];    int yp = y1 + diry[d1];    if(inmap(xp, yp) && !vis2[xp][yp])    {        xx2 = xp;        yy2 = yp;        cnty++;        cop[cnty].x = xx2;        cop[cnty].y = yy2;        DFSY(xx2, yy2, d1);    }    else    {        if(d1 == 0)        {            int tx = x1 + dirx[3];            int ty = y1 + diry[3];            if(!vis2[tx][ty] && inmap(tx, ty))            {                xx2 = tx;                yy2 = ty;                cnty++;                cop[cnty].x = xx2;                cop[cnty].y = yy2;                DFSY(xx2, yy2, 3);            }            else                fp2 = 1;        }        else        {            int tx = x1 + dirx[d1-1];            int ty = y1 + diry[d1-1];            if(!vis2[tx][ty] && inmap(tx, ty))            {                xx2 = tx;                yy2 = ty;                cnty++;                cop[cnty].x = xx2;                cop[cnty].y = yy2;                DFSY(xx2, yy2, d1-1);            }            else                fp2 = 1;        }    }    if(fp2)        return;}int main(){    while(scanf("%d", &n))    {        if(!n)            break;        mem(vis1, 0);        mem(vis2, 0);        mem(num, 0);        cntx = 0;        cnty = 0;        scanf("%d%d%d", &st1x, &st1y, &dir1);        scanf("%d%d%d", &st2x, &st2y, &dir2);        fp1 = 0;        fp2 = 0;        xx1 = st1x;        yy1 = st1y;        xx2 = st2x;        yy2 = st2y;        flag = 0;        DFSX(st1x, st1y, dir1);        DFSY(st2x, st2y, dir2);        int inde = min(cntx, cnty);        int i;        for(i = 1; i < inde; ++i)        {            if(num[i].x == cop[i].x && num[i].y == cop[i].y)            {                flag = 1;                break;            }        }        if(flag)            printf("%d %d\n", num[i].x, num[i].y);        else            printf("-1\n");    }    return 0;}

 

 

后面就使用一步步判断的方法搞，及时跳出，这样就可以了。我又犯昨天的错误了，没有判断起点相同，WA了好几次：如下：

 

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;#define maxn 1001int vis1[maxn][maxn];int vis2[maxn][maxn];int n;int st1x, st1y, dir1;int st2x, st2y, dir2;int fp1;int fp2;int dirx[4] = {0, 1, 0, -1};int diry[4] = {1, 0, -1, 0};bool inmap(int x, int y){    if(x < 0 || x >= n || y < 0 || y >= n)        return false;    return true;}int xx1, yy1;int flag;int xx2, yy2;void work(int x1, int y1, int d1, int x2, int y2, int d2){    while(!fp1 || !fp2)    {        if(!fp1)        {            vis1[x1][y1] = 1;            int tx = x1 + dirx[d1];            int ty = y1 + diry[d1];            if(!vis1[tx][ty] && inmap(tx, ty))            {                x1 = tx;                y1 = ty;            }            else            {                if(d1 == 3)                {                    tx = x1 + dirx[0];                    ty = y1 + diry[0];                    if(inmap(tx, ty) && !vis1[tx][ty])                    {                        x1 = tx;                        y1 = ty;                        d1 = 0;                    }                    else                        fp1 = 1;                }                else                {                    tx = x1 + dirx[d1+1];                    ty = y1 + diry[d1+1];                    if(inmap(tx, ty) && !vis1[tx][ty])                    {                        x1 = tx;                        y1 = ty;                        d1 += 1;                    }                    else                        fp1 = 1;                }            }        }        if(!fp2)        {            vis2[x2][y2] = 1;            int tx = x2 + dirx[d2];            int ty = y2 + diry[d2];            if(!vis2[tx][ty] && inmap(tx, ty))            {                x2 = tx;                y2 = ty;            }            else            {                if(d2 == 0)                {                    tx = x2 + dirx[3];                    ty = y2 + diry[3];                    if(inmap(tx, ty) && !vis2[tx][ty])                    {                        x2 = tx;                        y2 = ty;                        d2 = 3;                    }                    else                        fp2 = 1;                }                else                {                    tx = x2 + dirx[d2-1];                    ty = y2 + diry[d2-1];                    if(inmap(tx, ty) && !vis2[tx][ty])                    {                        x2 = tx;                        y2 = ty;                        d2 -= 1;                    }                    else                        fp2 = 1;                }            }        }        if(x1 == x2 && y1 == y2)        {            flag = 1;            printf("%d %d\n", x1, y1);            break;        }    }}int main(){    while(scanf("%d", &n))    {        if(!n)            break;        mem(vis1, 0);        mem(vis2, 0);        scanf("%d%d%d", &st1x, &st1y, &dir1);        scanf("%d%d%d", &st2x, &st2y, &dir2);        fp1 = 0;        fp2 = 0;        xx1 = st1x;        yy1 = st1y;        xx2 = st2x;        yy2 = st2y;        flag = 0;        if(st1x == st2x && st1y == st2y)        {            printf("%d %d\n", st1x, st1y);            continue;        }        else        {            work(st1x, st1y, dir1, st2x, st2y, dir2);            if(!flag)                printf("-1\n");        }    }    return 0;}

唉，做了好久。。。。