• poj 1789 Truck History


    http://poj.org/problem?id=1789

      简单MST,要注意cnt和size清零!

    View Code
      1 #include <cstdio>
      2 #include <iostream>
      3 #include <cstring>
      4 
      5 using namespace std;
      6 const int MAXV = 2005, MAXE = 2005 * 2005, INF = (~0u)>>2;
      7 char car[MAXV][10];
      8 struct edge{
      9     int t, w, next;
     10 }es[MAXE];
     11 int h[MAXV], cnt, n, heap[MAXV], size, pos[MAXV], dist[MAXV];
     12 
     13 void init(){
     14     cnt = size = 0;
     15     memset(h, 0, sizeof(h));
     16     memset(heap, 0, sizeof(heap));
     17     memset(pos, 0, sizeof(pos));
     18     memset(dist, 0, sizeof(dist));
     19 }
     20 
     21 void addedge(int x, int y, int z){
     22     es[++cnt].t = y;
     23     es[cnt].next = h[x];
     24     es[cnt].w = z;
     25     h[x] = cnt;
     26 }
     27 
     28 void heapup(int k){
     29     while(k > 1){
     30         if(dist[heap[k >> 1]] > dist[heap[k]]){
     31             swap(pos[heap[k >> 1]], pos[heap[k]]);
     32             swap(heap[k >> 1], heap[k]);
     33             k >>= 1;
     34         }
     35         else break;
     36     }
     37 }
     38 void heapdown(int k){
     39     while((k << 1) <= size){
     40         int j;
     41         if((k << 1) == size || dist[heap[k << 1]] < dist[heap[k << 1 | 1]])
     42             j = k << 1;
     43         else
     44             j = k << 1 | 1;
     45         if(dist[heap[k]] > dist[heap[j]]){
     46             swap(pos[heap[k]], pos[heap[j]]);
     47             swap(heap[k], heap[j]);
     48             k = j;
     49         }
     50         else break;
     51     }
     52 }
     53 void push(int v, int d){
     54     dist[v] = d;
     55     heap[++size] = v;
     56     pos[v] = size;
     57     heapup(size);
     58 }
     59 
     60 int pop(){
     61     int ret = heap[1];
     62     swap(pos[heap[size]], pos[heap[1]]);
     63     swap(heap[size], heap[1]);
     64     size--;
     65     heapdown(1);
     66     return ret;
     67 }
     68 
     69 int prim(){
     70     int mst = 0, i, p;
     71 
     72     push(1, 0);
     73     for(i = 2; i <= n; i++)
     74         push(i, INF);
     75     for(i = 1; i <= n; i++){
     76         int t = pop();
     77         mst += dist[t];
     78         pos[t] = -1;
     79         for(p = h[t]; p; p = es[p].next){
     80             int dst = es[p].t;
     81             if(pos[dst] != -1 && dist[dst] > es[p].w){
     82                 dist[dst] = es[p].w;
     83                 heapup(pos[dst]);
     84                 heapdown(pos[dst]);
     85             }
     86         }
     87     }
     88     return mst;
     89 }
     90 
     91 int diff(char *a, char *b){
     92     int ret = 0;
     93 
     94     for (int i = 0; i < 7; i++){
     95         ret += (a[i] != b[i]);
     96     }
     97 
     98     return ret;
     99 }
    100 
    101 void deal(){
    102     int c;
    103 
    104     for (int i = 1; i <= n; i++){
    105         scanf("%s", car[i]);
    106         for (int j = 1; j < i; j++){
    107             c = diff(car[i], car[j]);
    108             addedge(i, j, c);
    109             addedge(j, i, c);
    110         }
    111     }
    112     printf("The highest possible quality is 1/%d.\n", prim());
    113 }
    114 
    115 int main(){
    116 #ifndef ONLINE_JUDGE
    117     freopen("in", "r", stdin);
    118 #endif
    119     while (~scanf("%d", &n) && n){
    120         init();
    121         deal();
    122     }
    123 
    124     return 0;
    125 }

    ——written by Lyon

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  • 原文地址:https://www.cnblogs.com/LyonLys/p/poj_1789_Lyon.html
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