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这题实质为裸的差分约束。
先看最短路模型:若d[v] >= d[u] + w, 则连边u->v,之后就变成了d[v] <= d[u] + w , 即d[v] – d[u] <= w。
再看题目给出的关系:b比a多的糖果数目不超过c个,即d[b] – d[a] <= c ,正好与上面模型一样,
所以连边a->b,最后用dij+heap求最短路就行啦。
ps:我用vector一直TLE,后来改用前向星才过了orz。。。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<queue> 5 #include<vector> 6 #include<set> 7 #define CLR(a,b) memset((a),(b),sizeof((a))) 8 using namespace std; 9 10 const int N = 30001; 11 const int M = 150001; 12 const int inf = 0x3f3f3f3f; 13 14 int n, m; 15 bool s[N]; 16 int head[N]; 17 int cnt; 18 struct edge{ 19 int nex; 20 int v, w; 21 }g[M]; 22 struct node{ 23 int v, w; 24 node(int _v=0,int _w=0):v(_v),w(_w){} 25 bool operator < (const node&r)const{ 26 return r.w < w; 27 } 28 }; 29 void add_edge(int u,int v,int w){ 30 g[cnt].v = v; 31 g[cnt].w = w; 32 g[cnt].nex = head[u]; 33 head[u] = cnt++; 34 } 35 void dij(){ 36 int i, u, v, w; 37 node t; 38 CLR(s, 0); 39 priority_queue<node>q; 40 q.push(node(1,0)); 41 while(!q.empty()){ 42 t = q.top(); q.pop(); 43 u = t.v; 44 if(s[u]) continue; 45 s[u] = 1; 46 if(u == n) break; 47 for(i = head[u]; ~i; i = g[i].nex){ 48 v = g[i].v; 49 if(!s[v]){ 50 w = t.w + g[i].w; 51 q.push(node(v, w)); 52 } 53 } 54 } 55 printf("%d\n", t.w); 56 } 57 int main(){ 58 int i, j, a, b, c; 59 scanf("%d %d", &n, &m); 60 cnt = 0; 61 CLR(head, -1); 62 for(i = 1; i <= m; ++i){ 63 scanf("%d %d %d", &a, &b, &c); 64 add_edge(a,b,c); 65 } 66 dij(); 67 return 0; 68 }
poj3159 Candies(差分约束,dij+heap)
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原文地址:http://www.cnblogs.com/GraceSkyer/p/5873520.html