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1003. Emergency (25)

2019-11-08 19:24:48
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As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.InputEach input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.OutputFor each test case, PRint in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.Sample Input5 6 0 21 2 1 5 30 1 10 2 20 3 11 2 12 4 13 4 1Sample Output2 4
//一開始/*amount[start] = team[start];dist[start] = 0;pathcount[start] = 1;u = 0;dist[1] = dist[0] + map[0][1]; // dist[1] == 1dist[2] = dist[0] + map[0][1]; // dist[2] == 2amount[1] = 1 + 2;pathcount[1] = 1;pathcount[2] = 1;u = 1;dist[1] + map[1][2] == 2;//滿足dist[2] == dist[1] _ map[1][2];pathcount[2] = 1 + 1;//此時amount[2] = 0,amount[1] == 3,team[2] == 1;//soamount[2] = 3 + 1;....此時0~2的最短距離和0~2的所有team數目已經求出來了,分別是pathcount[2],amount[2]*/
#include <cstdio>#include <iostream>#include <cstdlib>using namespace std;const int MX = 500 + 5;const int INF = 9999999;int map[MX][MX];//記錄兩個點之間的距離sint dist[MX];//記錄從起始點到點i的距離int teams[MX];//記錄每個城市的隊伍數量int amount[MX];//記錄起始點到當前點的最大隊伍數量int pathcount[MX];//記錄從起始點到當前點點的最短路徑數量int v[MX];//標記是否被訪問過int N,M,start,en,dmin;void dijkstra(int start) {	amount[start] = teams[start];	dist[start] = 0;	pathcount[start] = 1;	while (1){        int u,dmin = INF;        //都是從0開始,因為第一遍循環會把起始點找出來        for (int i = 0; i < N; i++){            if (v[i] == 0 && dist[i] < dmin){                dmin = dist[i];                u = i;            }        }		//dmin==INF表示所有點已經都遍歷過了        if (dmin == INF)            break;		//將找到的點標記一下,說明這個點已經訪問過,之后就不要訪問了		v[u] = 1;		for (int i = 0; i < N; i++) {			if (v[i] == 0) {				if (dist[i] > dist[u] + map[u][i]) {					dist[i] = dist[u] + map[u][i];					amount[i] = amount[u] + teams[i];					pathcount[i] = pathcount[u];				}				else  if (dist[i] == dist[u] + map[u][i]) {					//如果距離相等就,到達當前的最短路徑數就加1                    pathcount[i] += pathcount[u];                    //如果之前的最大隊伍數量小于現在的隊伍數量,則進行更新                    if (amount[i] < amount[u] + teams[i]){                        amount[i] = amount[u] + teams[i];                    }				}			}		}	}	}int main() {	while (~scanf("%d%d%d%d",&N,&M,&start,&en)) {		for (int i = 0; i < N; i++) {			cin >> teams[i];		}		//初始化map數組		for (int i = 0; i < N; i++) {			dist[i] = INF;			for (int j = 0; j < N; j++) {				map[i][j] = INF;			}		}		//讀入數據		for (int i = 0; i < M; i++) {			int c1,c2,L;			cin >> c1 >> c2 >> L;			map[c1][c2] = map[c2][c1] = L;		}		//對所有點進行松弛		dijkstra(start);				cout << pathcount[en] << " " << amount[en] << endl;	}	system("pause");	return 0;}
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