编译原理简单的LALR(1)分析表的构造

闲着无聊，，，，写了一个简单的LALR(1)分析表的构造，就当是复习了

#include <cstdio>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
/*
1、输入文法,对非终结符和终结符进行映射
2、求出非终结符的FIRST集
3、求项目集的闭包函数(向前搜索符)
4、求GO函数
5、LR(1)分析表
6、合并LR(1)分析表内的同心项目集
7、LALR(1)分析表
8、输入串分析
注意：
1、输入产生式不超过100条
2、'@'代表空
3、'$'代表增加产生式$->S，拓广文法
4、终结符和非终结符的个数都不超过20
5、项目集不超过100
*/
struct grammar{
    char vn, s[10];//产生式vn->s[10]
}g[100];
int g_num;//记录产生式的个数
map<char, int> vn_map;//对非终结符进行映射
map<char, int> vt_map;//对终结符进行映射
int vn_num, vt_num;//记录非终结符和终结符的个数
char vn[20], vt[20];//存储非终结符和终结符的值
int first[20][20];//fisrt[i][j]代表非终结符vn[i]的first集包含终结符vt[j]
struct project{
    grammar g;//产生式
    int k, pre[20];//空格点位置k，向前搜索符pre
};
vector<project> closure[100];//项目集
int closure_num;//项目集的个数
struct Edge{
    int u, v;
    int next;
    char ch;
}edge[1000];//项目集closure[u]由条件ch转换为项目集closure[v]
int go[100];//GO函数头指针
int edge_num;//转换条件数目
int belong[100];//合并项目集后的归属
int merge_num;//合并后项目集的个数
int lalr1_action[100][20], lalr1_goto[100][20];//LALR(1)分析表的action和goto表
/*
读入文法G
    控制输入格式
    扩广文法
    对非终结符和终结符进行映射
*/
void read() {
    printf("请输入文法G,'@'代表空\n");
    printf("直接输入'@'输入结束\n");
    g_num = 0;
    int i, j, len, flag;
    char ch, str[20];
    grammar temp;
    while( gets(str) ) {
        if( str[0] == '@' ) break;
        len = strlen(str);
        flag = 0;
        if( len < 4 ) flag = 1;
        else if( str[0] < 'A' || str[0] > 'Z' || str[1] != '-' || str[2] != '>' ) flag = 1;
        else {
            temp.vn = str[0];
            for(i = 3; i < len; i++) {
                if( str[i] == ' ' ) break;
                temp.s[i-3] = str[i];
            }
            temp.s[i-3] = '\0';
            if( i < len ) flag = 1;
        }
        if( flag ) {
            printf("产生式%s不符合文法规则，已忽略。\n", str);
            continue ;
        }
        else {
            g[ ++g_num ] = temp;
        }
    }
    if( g_num == 0 ) {
        printf("未成功输入产生式\n");
        return;
    }
    printf("输入完成，共输入%d条产生式\n", g_num);
    g[0].vn = '$';
    g[0].s[0] = g[1].vn;
    g[0].s[1] = '\0';
    printf("拓广文法:%c->%s\n", g[0].vn, g[0].s);
    for(i = 1; i <= g_num; i++) {
        printf("\t %c->%s\n", g[i].vn, g[i].s);
    }
    vn_num = vt_num = 0;
    if( vt_map['@'] == 0 ) {
        vt_map['@'] = ++vt_num;
        vt[vt_num] = '@';
    }
    for(i = 0; i <= g_num; i++) {
        ch = g[i].vn;
        if( vn_map[ch] == 0 ) {
            vn_map[ch] = ++vn_num;
            vn[vn_num] = ch;
        }
        len = strlen( g[i].s );
        for(j = 0; j < len; j++) {
            ch = g[i].s[j];
            if( ch >= 'A' && ch <= 'Z' ) continue;
            if( vt_map[ch] == 0 ) {
                vt_map[ch] = ++vt_num;
                vt[vt_num] = ch;
            }
        }
    }
    if( vt_map['#'] == 0 ) {
        vt_map['#'] = ++vt_num;
        vt[vt_num] = '#';
    }
    for(i = 1, printf("非终结符:\n"); i <= vn_num; i++) {
        printf("%c -> %d\n", vn[i], vn_map[ vn[i] ]);
    }
    for(i = 1, printf("终结符:\n"); i <= vt_num; i++) {
        printf("%c -> %d\n", vt[i], vt_map[ vt[i] ]);
    }
    return;
}
/*
求first集
    solve_fisrt()求出所有的非终结符的first集
    dfs(char ch)求非终结符ch的first集
*/
void dfs(char ch,int acc[],int vis[],int val[]) {
    int i, j, k, c = vn_map[ch];
    if( acc[c] ) {
        for(i = 1; i <= vt_num; i++)
            val[i] = first[c][i];
        return;
    }
    int value[20];
    memset(value,0,sizeof(value));
    for(i = 0; i <= g_num; i++) {
        if( vis[i] ) continue;
        vis[i] = 1;
        if( g[i].vn != ch ) continue;
        int len = strlen( g[i].s );
        for(j = 0; j < len; j++) {
            char sh = g[i].s[j];
            if( vn_map[sh] ) {
                dfs(sh,acc,vis,value);
                for(k = 2; k <= vt_num; k++) {
                    if( value[k] ) val[k] = 1;
                }
                if( value[1] == 0 ) break;
            }
            else {
                c = vt_map[sh];
                val[c] = 1;
                break;
            }
        }
        if( j == len ) val[1] = 1;
    }
    return;
}
void solve_first() {
    int acc[20], vis[20], value[20];
    int i, j, k;
    memset(first,0,sizeof(first));
    memset(acc,0,sizeof(acc));
    for(i = 1; i <= vn_num; i++) {
        if( acc[i] ) continue;
        memset(vis,0,sizeof(vis));
        memset(value,0,sizeof(value));
        dfs(vn[i],acc,vis,value);
        for(j = 1; j <= vt_num; j++)
            first[i][j] = value[j];
        acc[i] = 1;
    }
    printf("输出first集\n");
    printf("\t");
    for(i = 1; i <= vt_num; i++)
        printf("%c\t", vt[i]);
    printf("\n");
    for(i = 1; i <= vn_num; i++) {
        printf("%c\t", vn[i]);
        for(j = 1; j <= vt_num; j++)
            printf("%d\t", first[i][j]);
        printf("\n");
    }
    return;
}
/*
求解闭包函数/go函数
    newproject()获得一个新的项目
    solve_closure()求项目集的闭包
    solve_projects()求所有的项目集
*/
int Equal(grammar u,grammar v) {
    if( u.vn != v.vn ) return 0;
    if( !strcmp(u.s,v.s) ) return 1;
    return 0;
}
int Equal(project u,project v) {
    if( !Equal(u.g,v.g) ) return 0;
    if( u.k != v.k ) return 0;
    for(int i = 1; i <= vt_num; i++)
        if( u.pre[i] != v.pre[i] ) return 0;
    return 1;
}
int Equal(int x,int y) {
    int i, j;
    if( closure[x].size() != closure[y].size() ) return 0;
    for(i = 0; i < closure[x].size(); i++) {
        project u = closure[x][i];
        for(j = 0; j < closure[y].size(); j++) {
            project v = closure[y][j];
            if( Equal(u,v) ) break;
        }
        if( j == closure[y].size() ) break;
    }
    if( i == closure[x].size() ) return 1;
    return 0;
}
project newproject() {
    project temp;
    memset(temp.pre,0,sizeof(temp.pre));
    return temp;
}
int addproject(project temp,int c) {
    for(int i = 0; i < closure[c].size(); i++) {
        project old = closure[c][i];
        if( Equal(old.g,temp.g) && old.k == temp.k ) return i;
    }
    return -1;
}
int solve_closure(int c){
    int i, j, k, l;
    for(i = 0; i < closure[c].size(); i++) {
        project old = closure[c][i];
        int len = strlen( old.g.s );
        if( old.k == len ) continue;
        char vn = old.g.s[ old.k ];
        if( vt_map[vn] ) break;
        for(j = 0; j <= g_num; j++) {
            if( g[j].vn != vn ) continue;
            //得到新添加项目temp
            project temp = newproject();
            temp.g = g[j];
            temp.k = 0;
            //排除S->.@的情况
            for(k = 0; k < strlen( g[j].s ); k++) {
                if( g[j].s[k] == '@' ) temp.k++;
                else break;
            }
            char ch = old.g.s[old.k+1];
            if( ch == '\0' ) {
                for(k = 1; k <= vt_num; k++)
                    temp.pre[k] = old.pre[k];
            }
            else if( vn_map[ch] ) {
                for(k = 1; k <= vt_num; k++)
                    temp.pre[k] = first[ vn_map[ch] ][k];
            }
            else
                temp.pre[ vt_map[ch] ] = 1;
            //判断temp是否已经存在在项目集中
            int flag = addproject(temp,c);
            if( flag == -1 ) closure[c].push_back(temp);
            else {
                for(k = 1; k <= vt_num; k++)
                    if( temp.pre[k] ) closure[c][flag].pre[k] = 1;
            }
        }
    }
    //完成新的项目集closure[c]，判断是否出现过
    for(i = 1; i < c; i++) {
        if( Equal(i,c) ) return i;
    }
    return c;
}
void addedge(int u,int v,char ch) {//设置GO函数的边
    edge[edge_num].u = u;
    edge[edge_num].v = v;
    edge[edge_num].ch = ch;
    edge[edge_num].next = go[u];
    go[u] = edge_num++;
}
void solve_projects() {
    project temp = newproject();
    //初始化项目集，仅包含g[0],向前搜索符包含'#'
    closure_num = 0;
    temp.g = g[0];
    temp.k = 0;
    temp.pre[ vt_map['#'] ] = 1;
    closure[closure_num+1].clear();
    closure[closure_num+1].push_back(temp);
    int c = solve_closure(closure_num+1);
    if( c == closure_num+1 ) {
        closure_num++;
    }
    int i, j, k;
    //求GO函数
    memset(go,-1,sizeof(go));
    edge_num = 0;
    for(i = 1; i <= closure_num; i++) {
        //测试项目集输出
        printf("项目集%d包含:\n", i);
        for(j = 0; j < closure[i].size(); j++) {
            project old = closure[i][j];
            printf("\t%c->", old.g.vn);
            for(k = 0; k <= strlen( old.g.s ); k++) {
                if( k == old.k ) printf(".");
                printf("%c", old.g.s[k]);
            }
            printf("\t");
            for(k = 1; k <= vt_num; k++)
                if( old.pre[k] ) printf("%c ", vt[k]);
            printf("\n");
        }
        //求GO函数
        int vis[100];
        memset(vis,0,sizeof(vis));
        for(j = 0; j < closure[i].size(); j++) {
            if( vis[j] ) continue;
            project old = closure[i][j];
            int len = strlen(old.g.s);
            if( old.k == len ) continue;
            closure[ closure_num+1 ].clear();
            for(k = j; k < closure[i].size(); k++) {
                project oldd = closure[i][k];
                if( oldd.g.s[ oldd.k ] == old.g.s[ old.k ] ) {
                    vis[k] = 1;
                    project temp = oldd;
                    temp.k++;
                    closure[ closure_num+1 ].push_back(temp);
                }
            }
            if( closure[ closure_num+1 ].size() == 0 ) continue;
            c = solve_closure(closure_num+1);
            if( c == closure_num+1 ) {
                closure_num++;
            }
            addedge(i,c,old.g.s[ old.k ]);
        }
    }
    return;
}
/*
输出LR(1)分析表
    所有状态从1到closure_num
    lr1_action中非负数i代表第i个产生式，负数-i代表状态i
    lr1_goto中负数-i代表状态i
*/
int findgrammar(grammar temp) {
    for(int i = 0; i <= g_num; i++) {
        if( Equal(temp,g[i]) ) return i;
    }
}
void printLR1() {
    int flag = 0;
    int lr1_action[100][20], lr1_goto[100][20];
    memset(lr1_action,INF,sizeof(lr1_action));
    memset(lr1_goto,INF,sizeof(lr1_goto));
    int i, j, k;
    for(i = 1; i <= closure_num; i++) {
        for(j = 0; j < closure[i].size(); j++) {
            project old = closure[i][j];
            char ch = old.g.s[ old.k ];
            int c = findgrammar(old.g);
            if( ch == '\0' ) {
                for(k = 1; k <= vt_num; k++) {
                    if( old.pre[k] ) {
                        if( lr1_action[i][k] == INF || lr1_action[i][k] == c )
                            lr1_action[i][k] = c;
                        else flag = 1;
                    }
                }
            }
        }
        for(j = go[i]; j != -1; j = edge[j].next) {
            int u = edge[j].u, v = edge[j].v;
            char ch = edge[j].ch;
            if( vn_map[ch] ) {
                if( lr1_goto[u][ vn_map[ch] ] == INF || lr1_goto[u][ vn_map[ch] ] == -v )
                    lr1_goto[u][ vn_map[ch] ] = -v;
                else flag = 1;
            }
            else {
                if( lr1_action[u][ vt_map[ch] ] == INF || lr1_action[u][ vt_map[ch] ] == -v )
                    lr1_action[u][ vt_map[ch] ] = -v;
                else flag = 1;
            }
        }
    }
    if( flag ) {
        printf("出现冲突，该文法不是LR(1)文法\n");
        return;
    }
    printf("LR(1)分析表：\n");
    printf("状态\t");
    for(i = 1; i <= vt_num; i++)
        printf("%c\t", vt[i]);
    for(i = 1; i <= vn_num; i++)
        printf("%c\t", vn[i]);
    printf("\n");
    for(i = 1; i <= closure_num; i++) {
        printf("%d\t", i);
        for(j = 1; j <= vt_num; j++) {
            k = lr1_action[i][j];
            if( k == INF ) printf("\t");
            else if( k < 0 ) printf("S%d\t", -k);
            else printf("r%d\t", k);
        }
        for(j = 1; j <= vn_num; j++) {
            k = lr1_goto[i][j];
            if( k == INF ) printf("\t");
            else if( k < 0 ) printf("%d\t", -k);
            else printf("r%d\t", k);
        }
        printf("\n");
    }
    return ;
}
/*
合并同心项目集
*/
int Equalproject(int x,int y) {
    if( closure[x].size() != closure[y].size() ) return 0;
    int i, j, k;
    for(i = 0; i < closure[x].size(); i++) {
        project u = closure[x][i];
        for(j = 0; j < closure[y].size(); j++) {
            project v = closure[y][j];
            if( u.k == v.k && Equal(u.g,v.g) ) break;
        }
        if( j == closure[y].size() ) return 0;
    }
    return 1;
}
void project_merge() {
    int i, j, k;
    merge_num = 0;//num表示合并后的状态编号索引
    int vis[100];
    memset(vis,0,sizeof(vis));
    for(i = 1; i <= closure_num; i++) {
        if( vis[i] ) continue;
        belong[i] = ++merge_num;
        vis[i] = 1;
        for(j = i+1; j <= closure_num; j++) {
            if( Equalproject(i,j) ) {
                belong[j] = merge_num;
                vis[j] = 1;
            }
        }
    }
    printf("项目集合并:\n");
    for(i = 1; i <= merge_num; i++) {
        printf("新项目集%d包含项目集:", i);
        for(j = 1; j <= closure_num; j++)
            if( belong[j] == i ) printf("%d ", j);
        printf("\n");
    }
}
/*
构造LALR(1)分析表
*/
void solve_lalr1() {
    int flag = 0;
    memset(lalr1_action,INF,sizeof(lalr1_action));
    memset(lalr1_goto,INF,sizeof(lalr1_goto));
    int i, j, k;
    for(i = 1; i <= closure_num; i++) {
        for(j = 0; j < closure[i].size(); j++) {
            project old = closure[i][j];
            char ch = old.g.s[ old.k ];
            int c = findgrammar(old.g);
            if( ch == '\0' ) {
                for(k = 1; k <= vt_num; k++) {
                    if( old.pre[k] ) {
                        if( lalr1_action[ belong[i] ][k] == INF || lalr1_action[ belong[i] ][k] == c )
                            lalr1_action[ belong[i] ][k] = c;
                        else flag = 1;
                    }
                }
            }
        }
        for(j = go[i]; j != -1; j = edge[j].next) {
            int u = belong[edge[j].u], v = belong[edge[j].v];
            char ch = edge[j].ch;
            if( vn_map[ch] ) {
                if( lalr1_goto[u][ vn_map[ch] ] == INF || lalr1_goto[u][ vn_map[ch] ] == -v )
                    lalr1_goto[u][ vn_map[ch] ] = -v;
                else flag = 1;
            }
            else {
                if( lalr1_action[u][ vt_map[ch] ] == INF || lalr1_action[u][ vt_map[ch] ] == -v )
                    lalr1_action[u][ vt_map[ch] ] = -v;
                else flag = 1;
            }
        }
    }
    if( flag ) {
        printf("出现冲突，该文法不是LALR(1)文法\n");
        return;
    }
    printf("LALR(1)分析表：\n");
    printf("状态\t");
    for(i = 1; i <= vt_num; i++)
        printf("%c\t", vt[i]);
    for(i = 1; i <= vn_num; i++)
        printf("%c\t", vn[i]);
    printf("\n");
    for(i = 1; i <= merge_num; i++) {
        printf("%d\t", i);
        for(j = 1; j <= vt_num; j++) {
            k = lalr1_action[i][j];
            if( k == INF ) printf("\t");
            else if( k < 0 ) printf("S%d\t", -k);
            else printf("r%d\t", k);
        }
        for(j = 1; j <= vn_num; j++) {
            k = lalr1_goto[i][j];
            if( k == INF ) printf("\t");
            else if( k < 0 ) printf("%d\t", -k);
            else printf("r%d\t", k);
        }
        printf("\n");
    }
    return ;
}
/*
对输入串进行匹配
    当产生式S->@，不需要符号栈退出字符
*/
void solve_str() {
    printf("请输入输入串：\n");
    char str[100];
    int step = 0, sta[100];
    char sym[100];
    int sta_num = 0, sym_num = 0;
    sta[sta_num++] = 1;
    sym[sym_num++] = '#';
    int i, j, k, len;
    scanf("%s", str);
    len = strlen(str);
    printf("步骤\t状态栈\t符号栈\t输入串\tACTION\tGOTO\t\n");
    k = 0;
    while(k < len) {
        printf("%d\t", ++step);
        for(i = 0; i < sta_num; i++)
            printf("%d", sta[i]);
        printf("\t");
        for(i = 0; i < sym_num; i++)
            printf("%c", sym[i]);
        printf("\t");
        for(i = k; i < len; i++)
            printf("%c", str[i]);
        printf("\t");
        if( sta_num == 0 ) {
            printf("状态栈为空，错误\n");
            break;
        }
        int x = sta[ sta_num-1 ];
        int y = vt_map[ str[k] ];
        int c = lalr1_action[x][y];
        if( c == INF ) {
            printf("ACTION函数为空,错误\n");
            break;
        }
        if( c < 0 ) {
            sta[ sta_num++ ] = -c;
            sym[ sym_num++ ] = str[k++];
            printf("S%d\t\t", -c);
        }
        else {
            if( c == 0 ) {
                printf("acc\n");
                break;
            }
            for(i = 0; i < strlen(g[c].s); i++) {
                if( g[c].s[i] != '@' ) {
                    sym_num--;
                    sta_num--;
                }
            }
            if( sym_num < 0 || sta_num < 0 ) {
                printf("归约失败\n");
                break;
            }
            sym[ sym_num++ ] = g[c].vn;
            x = sta[ sta_num-1 ];
            y = vn_map[ g[c].vn ];
            if( lalr1_goto[x][y] == INF ) {
                printf("GO函数为空，错误\n");
            }
            sta[ sta_num++ ] = -lalr1_goto[x][y];
            printf("r%d\t%d\t", c, -lalr1_goto[x][y]);
        }
        printf("\n");
    }
    return;
}
int main() {
    freopen("in2.txt","r",stdin);
    read();//读入文法G(S)
    solve_first();//求非终结符的first集
    solve_projects();//求闭包函数，go函数
    printLR1();//输出LR(1)分析表
    project_merge();//合并同心项目集
    solve_lalr1();//计算LALR(1)分析表
    solve_str();//输入串处理
    return 0;
}