poj 3159 Candies (dij + heap)

　　明明找的是差分约束，然后就找到这题不知道为什么是求1~n的最短路的题了。然后自己无聊写了一个heap，518ms通过。
代码如下：
  1 #include <cstdio>
  2 #include <cstring>
  3 #include <iostream>
  4 #include <algorithm>
  5 
  6 using namespace std;
  7 
  8 const int N = 33333;
  9 const int M = 155555;
 10 struct Edge {
 11     int t, nx, c;
 12 } edge[M];
 13 int eh[N], ec;
 14 
 15 void init() {
 16     memset(eh, -1, sizeof(eh));
 17     ec = 0;
 18 }
 19 
 20 void addedge(int s, int t, int c) {
 21     edge[ec].t = t, edge[ec].nx = eh[s], edge[ec].c = c;
 22     eh[s] = ec++;
 23 }
 24 
 25 struct Heap {
 26     int val[N], pos[N], id[N], sz, tmp, d;
 27     void init() {
 28         sz = 0;
 29         memset(pos, -1, sizeof(pos));
 30     }
 31     int up(int t) {
 32         while (t > 1 && val[t] < val[t >> 1]) pos[id[t] = id[t >> 1]] = t, swap(val[t], val[t >> 1]), t >>= 1;
 33         return t;
 34     }
 35     void push(int i, int v) {
 36         val[++sz] = v;
 37         pos[i] = up(sz);
 38         id[pos[i]] = i;
 39     }
 40     void down(int tmp) {
 41         while ((tmp << 1) <= sz) {
 42             if (val[tmp] <= val[tmp << 1] && val[tmp] <= val[tmp << 1 | 1]) break;
 43             if ((tmp << 1 == sz) || val[tmp << 1] <= val[tmp << 1 | 1]) {
 44                 d = tmp << 1;
 45                 swap(pos[id[tmp]], pos[id[d]]);
 46                 swap(id[tmp], id[d]);
 47                 swap(val[tmp], val[d]);
 48                 tmp = d;
 49             } else {
 50                 d = tmp << 1 | 1;
 51                 swap(pos[id[tmp]], pos[id[d]]);
 52                 swap(id[tmp], id[d]);
 53                 swap(val[tmp], val[d]);
 54                 tmp = d;
 55             }
 56         }
 57     }
 58     void modify(int i, int v) {
 59         val[pos[i]] = v;
 60         down(pos[i]);
 61         pos[i] = up(pos[i]);
 62         id[pos[i]] = i;
 63     }
 64     void pop(int &i) {
 65         if (sz == 0) {
 66             i = -1;
 67             return ;
 68         }
 69         i = id[1];
 70         pos[i] = -1;
 71         if (sz == 1) { sz--; return ;}
 72         val[1] = val[sz];
 73         id[1] = id[sz];
 74         pos[id[1]] = 1;
 75         sz--;
 76         if (sz > 0) down(1);
 77     }
 78     int gv(int i) { return ~pos[i] ? val[pos[i]] : -1;}
 79 } hp;
 80 int dis[N];
 81 
 82 int dij(int s, int t) {
 83     int cur, i;
 84     hp.init();
 85     memset(dis, 127, sizeof(dis));
 86     dis[s] = 0;
 87     hp.push(s, 0);
 88     while (hp.sz) {
 89         hp.pop(cur);
 90         if (cur == t) return dis[cur];
 91         for (i = eh[cur]; ~i; i = edge[i].nx) {
 92             Edge &e = edge[i];
 93             if (dis[e.t] > dis[cur] + e.c) {
 94                 dis[e.t] = dis[cur] + e.c;
 95                 if (~hp.gv(e.t)) hp.modify(e.t, dis[e.t]);
 96                 else hp.push(e.t, dis[e.t]);
 97             }
 98         }
 99     }
100     return -1;
101 }
102 
103 int main() {
104     int x, y, v;
105     int n, m;
106     while (~scanf("%d%d", &n, &m)) {
107         init();
108         while (m--) {
109             scanf("%d%d%d", &x, &y, &v);
110             addedge(x, y, v);
111         }
112         printf("%d\n", dij(1, n));
113     }
114     return 0;
115 }
View Code
——written by Lyon
posted @ 2013-07-21 22:16 LyonLys 阅读(243) 评论(0) 收藏举报
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