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思路： 建图！！！ 之后枚举最大等级，求到物品1的最短路。
1 2 #include <stdio.h> 3 #include <queue> 4 #include <string.h> 5 #define maxn 0x3ffffff 6 using namespace std; 7 typedef pair<int,int> pii; 8 int visit[110],dis[110],map[110][110],n,m,v,l,p,x,t,vis[110],liv[110],sb,w,cas; 9 priority_queue<pii,vector<pii>,greater<pii> >qq; 10 int main() 11 { 12 int i,j,max; 13 while(scanf("%d%d",&m,&n)){ 14 if(n==0&&m==0) 15 break; 16 memset(liv,0,sizeof(liv)); 17 memset(map,9,sizeof(map)); 18 sb=1; 19 max=0; 20 int sbsbsb=n; 21 while(sbsbsb--) 22 { 23 scanf("%d%d%d",&p,&l,&x); 24 map[0][sb]=p; 25 liv[sb]=l; 26 if(liv[sb]>max) 27 max=liv[sb]; 28 for(i=1;i<=x;i++) 29 { 30 scanf("%d%d",&t,&v); 31 map[t][sb]=v; 32 } 33 sb++; 34 } 35 int min=maxn; 36 for(i=0;i<=max;i++) 37 { 38 memset(visit,0,sizeof(visit)); 39 memset(dis,9,sizeof(dis)); 40 dis[0]=0; 41 qq.push(make_pair(dis[0],0)); 42 while(!qq.empty()) 43 { 44 pii u=qq.top(); 45 qq.pop(); 46 int x=u.second; 47 if(visit[x]) 48 continue; 49 visit[x]=1; 50 for(j=0;j<=n;j++) 51 { 52 if(liv[j]>=(i-m)&&liv[j]<=i) 53 { 54 if(dis[j]>(dis[x]+map[x][j])) 55 { 56 dis[j]=dis[x]+map[x][j]; 57 qq.push(make_pair(dis[j],j)); 58 } 59 } 60 } 61 } 62 if(dis[1]<min) 63 min=dis[1]; 64 } 65 printf("%d\n",min); 66 } 67 return 0; 68 }
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<map> #include<iomanip> #include<climits> #include<string.h> #include<cmath> #include<stdlib.h> #include<vector> #include<set> #define INF 1e7 #define MAXN 100010 #define maxn 111 #define maxm 1000 #define Mod 1000007 #define MIN(a,b) (a < b ? a : b) #define MAX(a,b) (a > b ? a : b) #define mem(a) memset(a,0,sizeof(a)) using namespace std; typedef long long LL; int n, m, p, l, x; int G[111][111], price[111], level[111]; int vis[111], dis[111]; int ans, k, t, v; void init() { memset(price,0,sizeof(price)); memset(level,0,sizeof(level)); for (int i = 0; i <= m; ++i) for (int j = 0; j <= m; ++j) { G[i][j] = INF; } } int Dijkstra() { int pos, min; for (int i = 1; i <= m; ++i) dis[i] = price[i]; for (int i = 1; i <= m; ++i) { min = INF; for (int j = 1; j <= m; ++j) { if (!vis[j] && min > dis[j]) { pos = j; min = dis[j]; } } vis[pos] = 1; for (int j = 1; j <= m; ++j) if (!vis[j] && dis[pos] + G[pos][j] < dis[j]) dis[j] = dis[pos] + G[pos][j]; } return ans = MIN(ans,dis[1]); } void deal() { ans = INF; for (int i = 1; i <= m; ++i) { int tmp = level[i]; for (int j = 1; j <= m; ++j) { if (level[j] - tmp > n || tmp > level[j]) vis[j] = 1; else vis[j] = 0; } Dijkstra(); } printf("%d\n",ans); } void read() { for (int i = 1; i <= m; ++i) { scanf("%d%d%d", &price[i], &level[i], &k); for (int j = 0; j < k; ++j) { scanf("%d%d", &t, &v); G[t][i] = v; } } } void run() { init(); read(); deal(); } int main() { while (~scanf("%d%d", &n, &m),n+m) run(); return 0; }
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原文地址：http://www.cnblogs.com/usedrosee/p/4338909.html
