poj 1062 昂贵的聘礼

http://162.105.81.212/JudgeOnline/problem?id=1062

一个特别要注意的就是题目里描述的"等级制度", 可以枚举

for(j=1; j<=n; j++) if(lev[j] >= lev[1] - m + i && lev[j] <= lev[1] + i)//枚举等级允许范围的结点 limit[j] = 1;

最小花费用的dijkstra算法求的, 在初始化dist[]数组时要注意,是否允许他们交易

for(i=1; i<=n; i++) { if(limit[i]) dist[i] = map[1][i]; else dist[i] = INF; visted[i] = 0; }

dijkstra算法都好久没写了, 照着书本写的- -

#include<iostream> using namespace std; #define INF 0x7ffffff #define MAX 101 int map[MAX][MAX], value[MAX], lev[MAX], dist[MAX]; int m, n; bool limit[MAX], visted[MAX]; //dijkstra求最短路径, dist[i]表示始点(1)到i点的花费 int dijkstra() { int i, j, k, mindist; for(i=1; i<=n; i++) { if(limit[i]) dist[i] = map[1][i]; else dist[i] = INF; visted[i] = 0; } dist[1] = 0; for(i=1; i<=n; i++) { mindist = INF; //选取最小的dist值 for(j=1; j<=n; j++) { if(!visted[j] && limit[j] && dist[j] < mindist) { k = j; mindist = dist[j]; } } visted[k] = 1; //更新所有未访问节点的dist值 for(j=1; j<=n; j++) { if(!visted[j] && limit[j]) if(map[k][j] < INF && dist[k] + map[k][j] < dist[j]) dist[j] = dist[k] + map[k][j]; } } mindist = INF; for(i=1; i<=n; i++) { dist[i] += value[i]; if(dist[i] < mindist) mindist = dist[i]; } return mindist; } int main() { int i, j, t, tmp, min; while(scanf("%d%d", &m, &n) != EOF) { //构图 for(i=1; i<=n; i++) for(j=1; j<=n; j++) map[i][j] = i==j?0:INF; for(i=1; i<=n; i++) { scanf("%d%d%d", &value[i], &lev[i], &t); for(j=1; j<=t; j++) { scanf("%d", &tmp); scanf("%d", &map[i][tmp]); } } min = value[1]; for(i=0; i<=m; i++) { memset(limit, 0, sizeof(limit)); for(j=1; j<=n; j++) if(lev[j] >= lev[1] - m + i && lev[j] <= lev[1] + i)//枚举等级允许范围的结点 limit[j] = 1; tmp = dijkstra(); if(min > tmp) min = tmp; } printf("%d/n", min); } return 0; }