uva 140

就是一道 DFS剪枝+枚举全排列 的题目
我用了vector来存点对的，不用考虑越界的问题。不过效率貌似不如直接用数组高。
WA了三次，花了好长时间找原因╮(╯▽╰)╭
多么弱智的错误啊=。=
WA代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <math.h>
#include <map>
#include <set>
using namespace std;

vector<vector<int> > align;
vector<int> dict;
set<int> s;
int n;
int A[27];
int B[27];
int ans; 

void parse(string cmd){
    int len = cmd.length();
    int f = 1;
    int x = -1;
    for(int i=0;i<len;i++){
        if(cmd[i]>=‘A‘ && cmd[i] <=‘Z‘){
            if(!s.count(cmd[i]-‘A‘)) {
            dict.push_back(cmd[i]-‘A‘);s.insert(cmd[i]-‘A‘);}
        }
        if(cmd[i] == ‘ ‘) continue;
        if(cmd[i] == ‘:‘) {f = 0;continue;}
        if(cmd[i] == ‘;‘) {f = 1;continue;}
        if(f==1) x = cmd[i] - ‘A‘;
        vector<int> tmp;
        if(f==0){
            tmp.push_back(x);tmp.push_back(cmd[i]-‘A‘);
            align.push_back(tmp);
        }
    }
}

void dfs(int cur,int dist){
    if(cur==n ){
        if(dist < ans) {
            ans = dist;
            for(int i=0;i<n;i++) B[i] = A[i];
        }
        return;
    }
    //else if(dist > ans) return;
    else for(int i=0;i<n;i++){
        int ok = 1;
        //这里的循环不能在一起写，否则就会出现 A B B ** 也符合的情况，但这样竟然能过样例==
        for(int j = 0;j<cur;j++){
            if(A[j] == dict[i]) {ok = 0;break;}
            for(int k=0;k<align.size();k++){
                if((align[k][0] == dict[i] && align[k][1]==A[j])||(align[k][0]==A[j] && align[k][1]==dict[i])){
                    if(cur-j>=ans){ok=0;break;
                    if(cur-j>dist) dist = cur-j;
                }
            }
            if(!ok) break;
        }
        if(ok){
            A[cur] = dict[i];
            dfs(cur+1,dist);
        }
    }
}

int main(){
    string line;
    while(cin>>line){
        align.clear();dict.clear();s.clear(); 
        ans = 100000000;
        if(line[0]==‘#‘) break;
        parse(line);
        n = dict.size();
        sort(dict.begin(),dict.end());
        dfs(0,0);
        for(int i=0;i<n;i++){
        printf("%c ",‘A‘+B[i]);
        }
        printf("-> %d\n",ans);
    }
    return 0;
}

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <math.h>
#include <map>
#include <set>
using namespace std;

vector<vector<int> > align;
vector<int> dict;
set<int> s;
int n;
int A[270];
int B[270];
int ans; 

void parse(string cmd){
    int len = cmd.length();
    int f = 1;
    int x = -1;
    for(int i=0;i<len;i++){
        if(cmd[i]>=‘A‘ && cmd[i] <=‘Z‘){
            if(!s.count(cmd[i]-‘A‘)) {
            dict.push_back(cmd[i]-‘A‘);s.insert(cmd[i]-‘A‘);}
        }
        if(cmd[i] == ‘ ‘) continue;
        if(cmd[i] == ‘:‘) {f = 0;continue;}
        if(cmd[i] == ‘;‘) {f = 1;continue;}
        if(f==1) x = cmd[i] - ‘A‘;
        vector<int> tmp;
        if(f==0){
            tmp.push_back(x);tmp.push_back(cmd[i]-‘A‘);
            align.push_back(tmp);
        }
    }
}

void dfs(int cur,int dist){
    if(cur==n ){
        if(dist < ans) {
            ans = dist;
            for(int i=0;i<n;i++) B[i] = A[i];
            /*printf("ans:%d\n",ans);
            for(int i=0;i<n;i++) printf("%c ",A[i]+‘A‘);
            printf("\n");*/
        }
        return;
    }
    //else if(dist > ans) return;
    else for(int i=0;i<n;i++){
        int ok = 1;
        for(int j = 0;j<cur;j++){
            if(A[j] == dict[i]) {ok = 0;break;}
        }
        if(ok){
            for(int j = 0;j<cur;j++){
                for(int k=0;k<align.size();k++){
                if((align[k][0] == dict[i] && align[k][1]==A[j])||(align[k][0]==A[j] && align[k][1]==dict[i])){
                    if(cur-j>=ans){ok=0;break;} 
                    /*
                    for(int m =0;m<cur;m++) printf("%c ",A[m]+‘A‘);
                    printf("\nnow\n");
                    printf("%c %c\n",A[j]+‘A‘,dict[i]+‘A‘);*/
                    //cout<<dict[i]<<"  "<<A[j]<<"  "<<cur-j<<endl;
                    if(cur-j>dist){dist = cur-j;
                    /*printf("now\n");
                      printf("cur:%d i:%d j:%d %c %c ans:%d\n",cur,i,j,A[j]+‘A‘,dict[i]+‘A‘,dist);*/
                      break;}
                }
            }
                if(!ok) break;
            }
        }
        if(ok){
            A[cur] = dict[i];
            dfs(cur+1,dist);
        }
    }
}

int main(){
    string line;
    while(cin>>line){
        align.clear();dict.clear();s.clear(); 
        ans = 999999;
        if(line[0]==‘#‘) break;
        parse(line);
        n = dict.size();
        for(int i=0;i<n;i++) ;
        sort(dict.begin(),dict.end());
        dfs(0,0);
        for(int i=0;i<n;i++){
        printf("%c ",‘A‘+B[i]);
        }
        printf("-> %d\n",ans);
    }
    return 0;
}