PAT (Advanced Level) Practice - 1003 Emergency

1003 Emergency

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.

Input Specification:

Each 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 C​2 - 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, c​2 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.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between C1​​ and C​2, 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 Input:

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output:

2 4

分析

单源最短路径问题，且需要注意的是对比的key不只是distance还有对比the number of rescue teams，还有就是需要统计最短路径的数量。

代码实现

#include <stdio.h>
#include <stdlib.h>

#define MAXV 505
#define INF 10000000

typedef struct {
	int Nv, Ne;
	int Dist[MAXV][MAXV];
	int Visited[MAXV];
	int MinDist[MAXV];
	int Team[MAXV], MaxTeam[MAXV];
	int Path[MAXV];
	int Count[MAXV];
} Graph;

void InitializeG(Graph *ptrG) {
	int i, j;

	ptrG->Nv = ptrG->Ne = 0;
	for (i = 0; i < MAXV; i++) {
		for (j = 0; j < MAXV; j++) {
			ptrG->Dist[i][j] = INF;
		}
		ptrG->MinDist[i] = INF;
		ptrG->Visited[i] = 0;
		ptrG->Team[i] = ptrG->MaxTeam[i] = 0;
		ptrG->Path[i] = -1;
		ptrG->Count[i] = 0;
	}
}

void ReadG(Graph *ptrG, int *S, int *D) {
	int i, V1, V2, W, N, M;

	scanf("%d %d %d %d", &N, &M, S, D);
	ptrG->Nv = N; ptrG->Ne = M;
	for (i = 0; i < N; i++)
		scanf("%d", &(ptrG->Team[i]));
	for (i = 0; i < M; i++) {
		scanf("%d %d %d", &V1, &V2, &W);
		ptrG->Dist[V1][V2] = ptrG->Dist[V2][V1] = W;
	}
}

void Dijkstra(Graph *ptrG, int S) {
	int i, j, minD, minV, MinDist, MaxTeam;

	ptrG->MinDist[S] = 0;
	ptrG->MaxTeam[S] = ptrG->Team[S];
	ptrG->Count[S] = 1;

	for (i = 0; i < ptrG->Nv; i++) {
		minD = INF;
		for (j = 0; j < ptrG->Nv; j++)
			if (!ptrG->Visited[j] && minD > ptrG->MinDist[j]) {
				minD = ptrG->MinDist[j];
				minV = j;
			}
		if (minD < INF)
			ptrG->Visited[minV] = 1;
		else//图不连通 
			break;

		for (j = 0; j < ptrG->Nv; j++) {
			if (!ptrG->Visited[j]) {
				int sumDist = ptrG->MinDist[minV] + ptrG->Dist[minV][j];
				int sumTeam = ptrG->MaxTeam[minV] + ptrG->Team[j];
				if (sumDist < ptrG->MinDist[j]) {
					ptrG->Count[j] = ptrG->Count[minV];//发现有更短路径：最短路径条数等于该路径上前驱结点的最短路径条数 
					ptrG->MinDist[j] = sumDist;
					ptrG->MaxTeam[j] = sumTeam;
					ptrG->Path[j] = minV;
				}
				else if ((sumDist == ptrG->MinDist[j])) {//距离相等
					if (ptrG->MaxTeam[j] < sumTeam) {//但救援队更多 
						ptrG->MaxTeam[j] = sumTeam;
						ptrG->Path[j] = minV;
					}
					ptrG->Count[j] += ptrG->Count[minV];//发现等长路径：最短路径条数等于当前条数加上前驱结点的最短路径条数 	
				}
			}
		}
	}
}

void PrintPath(int path[], int S) {
	if (path[S] >= 0) {
		PrintPath(path, path[S]);
		printf(" %d", S);
	}
	else
		printf("%d", S);
}

int main(void) {
	int S, D, Count = 0;
	Graph G;
	InitializeG(&G);
	ReadG(&G, &S, &D);
	Dijkstra(&G, S);
	printf("%d %d\n", G.Count[D], G.MaxTeam[D]);
	//PrintPath(G.Path, D);

	return 0;
}

更多资料

英文原题网址、中文题目网址