UVA 140 Brandwidth 带宽 (dfs回溯)

看到next_permutation好像也能过╮(╯▽╰)╭

这题学习点:

1.建图做映射

2.通过定序枚举保证字典序最小

3.strtok,sscanf,strchr等函数又复习了一遍,尽管程序中没有实际用上

4.剪枝,或者回溯

#include<bits/stdc++.h>
using namespace std;

int G[8][8],deg[8];
bool vG[8][8];//判读连通
int pos[8];
bool vis[8];
int k;
int best[8];
int cnt;
int ID[26];
char rev_ID[8];

void dfs(int d,int width)
{
    if(d == cnt){
        if(width < k){
            k = width;
            memcpy(best,pos,sizeof(pos));
        }
        return;
    }
    for(int i = 0;i < cnt; i++) if(!vis[i]){//把i放在d位置
        pos[d] = i;

        //prune 计算i和之前确定位置的结点的最大距离
        int m = 0;
        for(int j = d-1; j >=0; j--) if(vG[i][pos[j]]) {
            m = max(m,d-j);
            if(m >= k) continue;//这题数据水了,这里写成return还是能过
        }

        //prune2 计算u和未确定位置的相邻点的个数,适用于简单图
        int ct = 0;
        for(int j = 0; j < deg[i]; j++) if(!vis[G[i][j]]) ct++;
        if(ct >= k) continue;

        vis[i] = 1;
        dfs(d+1,max(width,m));
        vis[i] = 0;
    }
}


int main()
{
   // freopen("in.txt","r",stdin);
    char buf[100];
    const int INF = 0x3fffffff;
    while(fgets(buf,100,stdin) && *buf!='#'){
        cnt = 0;
        memset(vG,0,sizeof(vG));
        memset(deg,0,sizeof(deg));
        memset(ID,-1,sizeof(ID));
        memset(vis,0,sizeof(vis));
        bool ap[26] ;
        memset(ap,0,sizeof(ap));
        for(char *cur = buf; *cur ; cur++) {
            int u = *cur - 'A';
            if(0<= u && u < 26 ) ap[u] = 1;
        }

        for(int i = 0; i < 26; i++){
            if(ap[i]){
                rev_ID[cnt] = i+'A';
                ID[i] = cnt++;
            }
        }

        for(char *cur = buf; *cur ; cur++) {
            int u = ID[*cur - 'A'];
            for(cur+=2 ; *cur != '
' && *cur != ';' ; cur++) {
                int v = ID[*cur - 'A'];
                vG[u][v] = vG[v][u] = 1;
            }
        }

        for(int i = 0; i < cnt; i++)
            for(int j = 0; j < cnt; j++){
                if(vG[i][j]) G[i][deg[i]++] = j;
            }
        k = INF;
        dfs(0,0);
        for(int i = 0; i < cnt; i++){
            printf("%c ",rev_ID[best[i]]);
        }
        printf("-> %d
",k);
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/jerryRey/p/4628726.html