poj1062 昂贵的聘礼 (dijkstra最短路算法)

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POJ 1062 昂贵的聘礼

题意: 大致意思就是寻找一条路,使得到达编号为1的点的总代价最少且这条路上的任意两点间的等级差不超过m, 每个点有各自的权值, 每条边(i, j)的权值指的是可替换j的权值;

解法: 取任一点假设为最优路中等级最低的点, 将<该等级的点以及与该等级等级差超过m的点去掉; 然后进行dijkstra最短路算法, 取剩下点中权值最小的点进行遍历

code:

/***************Problem from :1062Problem describe :data:****************/#include<iostream>#include<cstdlib>#include<cstdio>#include<algorithm>#include<cmath>#include<map>#include<stack>#include<queue>#include<ctime>#include<cstring>#include<vector>#include<string>#define ll __int64using namespace std;const int inf = 0x3f3f3f3f;const int maxn=105;int min_p;int p[maxn], edge[maxn][maxn], vis[maxn] ,L[maxn], d[maxn], x[maxn];void init(){memset(edge, -1, sizeof(edge));memset(L, 0, sizeof(L));memset(p, -1, sizeof(p));min_p=inf;return ;}int min(int a, int b){return (a<b)?a:b;}int n, m ;void dijkstra() {int i, j, k;for(i=1; i<=n; i++) d[i] = p[i]; //将所有的点的d赋值为该点的权值 while(true){int v=-1, MIN=inf;for(i=1; i<=n; i++){if(!vis[i] && d[i]<MIN){MIN = d[v = i];}}if(v==-1) break;vis[v]=1;for(i=1; i<=n; i++){if(!vis[i] && edge[v][i]!=-1){d[i] = min(d[v]+edge[v][i], d[i]);}}}min_p = min(min_p, d[1]);return ;}int main(){int i, j, k, id, pi;//freopen("in.txt","r",stdin);while(~scanf("%d%d", &m, &n)){init();for(i=1; i<=n; i++){scanf("%d %d %d", &p[i], &L[i], &x[i]);for(j=1; j<=x[i]; j++){scanf("%d %d", &id, &pi);edge[id][i] = (edge[id][i]<0)?pi:min(edge[id][i], pi);}}for(i=1; i<=n; i++){if(L[i]<L[1]-m || L[i]>L[1]+m) continue; //与L[1]不满足条件的点必定不会出现在最优路中 for(j=1; j<=n; j++){if( L[j]<L[i]-m || L[j]>L[i] ) vis[j] = 1; //设L[i]为等级最小的那个点   去掉不符合条件的点 else vis[j]=0;}dijkstra();}printf("%d\n", min_p);}return 0;}

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