最短路径:Dijkstra和Floyd

		for (int v = 0; v < n; v++) {
			if (vis[v] == false && G[u][v] != INF) {
				if (d[u] + G[u][v] < d[v]) {
					d[v] = d[u] + G[u][v];
					c[v] = c[u] + cost[u][v];
				}
				else if ((d[u] + G[u][v] == d[v] && c[u] + cost[u][v] < c[v]) {
					c[v]=c[u]+cost[u][v];
				}
			}
		}

给每个点增加一个点权

比如说每个城市收集到的物资，在最短路径有多条时选择收集物资最多的

用weight[u]表示城市u中的物资数目，并增加一个数组w[]，令从起点s到达顶点u可以收集到的最大物资为w[u]，初始化时只有w[s]为weight[s]，其余w[u]均为0，d[u[+G[u][v]

而当d[u]+G[u][v]==d[v]时，且w[u]+weight[v]>w[v]时更新w[v]： 

		for (int v = 0; v < n; v++) {
			if (vis[v] == false && G[u][v] != INF) {
				if (d[u] + G[u][v] < d[v]) {
					d[v] = d[u] + G[u][v];
					w[v] = w[u] + weight[v];
				}
				else if ((d[u] + G[u][v] == d[v] && w[u] + weight[v] > w[v]) {
					w[v]=w[u]+weight[v];
				}
			}
		}

直接问有多少条最短路径

只需要增加一个数组num[]，令从起点s到达顶点u的最短路径条数为num[u]，初始化时只有nums[s]为1，其余num[u]为0，这样就可以在d[u]+G[u][v]

而当d[u]+G[u][v]==d[v]时，将num[u]加到num[v]上：

		for (int v = 0; v < n; v++) {
			if (vis[v] == false && G[u][v] != INF) {
				if (d[u] + G[u][v] < d[v]) {
					d[v] = d[u] + G[u][v];
					num[v] = num[u];
				}
				else if (d[u] + G[u][v] == d[v]) {
					num[v] += num[u];
				}
			}
		}

习题

PAT A1003 Emergency

题目详情 - 1003 Emergency (pintia.cn)

1003 Emergency

分数 25

作者 CHEN, Yue 单位 浙江大学

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 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​.

Output Specification:

For 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 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

代码长度限制 16 KB

时间限制 400 ms

内存限制 64 MB

题意：给N个城市，M条无向边。每个城市有一定数目的救援小组，所有边的边权已知。现在给出起点和重点，求从起点到终点的最短路径条数以及最短路径上救援小组的数目之和，如果有多条最短路径，则输出数目之和最大的

#include 
 
using namespace std;
 
const int INF = 100000000;
const int MAXV = 505;
 
int n, m, st, ed;
int G[MAXV][MAXV], weight[MAXV];
bool vis[MAXV] = { false };
int d[MAXV], w[MAXV], num[MAXV];
 
void Dijkstra(int s) {
	fill(d, d + MAXV, INF);
	fill(num, num + MAXV, 0);
	fill(w, w + MAXV, 0);
	d[s] = 0;
	w[s] = weight[s];
	num[s] = 1;
	for (int i = 0; i < n; i++) {
		int u = -1, MIN = INF;
		for (int j = 0; j < n; j++) {
			if (vis[j] == false && d[j] < MIN) {
				u = j;
				MIN = d[j];
			}
		}
		if (u == -1)return;
		vis[u] = true;
		for (int v = 0; v < n; v++) {
			if (vis[v] == false && G[u][v]!=INF) {
				if (d[u] + G[u][v] < d[v]) {
					d[v] = d[u] + G[u][v];
					w[v] = w[u] + weight[v];
					num[v] = num[u];
				}
				else if (d[u] + G[u][v] == d[v]) {
					if (w[u] + weight[v] > w[v]) {
						w[v] = w[u] + weight[v];
					}
					num[v] += num[u];
				}
			}
		}
	}
}
 
int main() {
	cin >> n >> m >> st >> ed;
	for (int i = 0; i < n; i++) {
		cin >> weight[i];
	}
	fill(G[0], G[0] + MAXV * MAXV, INF);
	for (int i = 0; i < m; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		G[u][v] = G[v][u] = w;
	}
	Dijkstra(st);
	cout << num[ed] << " " << w[ed];
 
	return 0;
}

PAT A1030 Travel Plan

1030 Travel Plan

分数 30

作者 CHEN, Yue  单位 浙江大学

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤500) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input:

4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output:

0 2 3 3 40

代码长度限制 16 KB

时间限制 400 ms

内存限制 64 MB

题意：有N个城市，M条道路，并给出M条道路的距离属性和花费，给定起点S和目的地D，求从起点S到D的最短路径、最短距离和花费，如果有多条最短路径，选择花费最小的那条

#include 	
 
using namespace std;
 
const int INF = 100000000;
const int MAXV = 505;
 
int n, m, st, ed;
int G[MAXV][MAXV], cost[MAXV][MAXV];
int d[MAXV], c[MAXV], pre[MAXV];
bool vis[MAXV] = { false };
 
void Dijkstra(int s) {
	fill(d, d + MAXV, INF);
	fill(c, c + MAXV, 0);
	for (int i = 0; i < n; i++)pre[i] = i;
 
	d[s] = 0;
	c[s] = 0;
	for (int i = 0; i < n; i++) {
		int u = -1, MIN = INF;
		for (int j = 0; j < n; j++) {
			if (vis[j] == false && d[j] < MIN) {
				u = j;
				MIN = d[j];
			}
		}
 
		if (u == -1)return;
		vis[u] = true;
		for (int v = 0; v < n; v++) {
			if (vis[v] == false && G[u][v] != INF) {
				if (d[u] + G[u][v] < d[v]) {
					d[v] = d[u] + G[u][v];
					c[v] = c[u] + cost[u][v];
					pre[v] = u;
				}
				else if (d[u] + G[u][v] == d[v] && c[u] + cost[u][v] < c[v]) {
					c[v] = c[u] + cost[u][v];
					pre[v] = u;
				}
			}
		}
	}
}
 
void dfs(int v) {
	if (v == st) {
		cout << v << " ";
		return;
	}
	dfs(pre[v]);
	cout << v << " ";
}
 
int main() {
 
	cin >> n >> m >> st >> ed;
	fill(G[0], G[0] + MAXV * MAXV, INF);
	for (int i = 0; i < m; i++) {
		int c1, c2, dis, co;
		cin >> c1 >> c2 >> dis >> co;
		G[c1][c2] = G[c2][c1] = dis;
		cost[c1][c2] = cost[c2][c1] = co;
	}
	Dijkstra(st);
	dfs(ed);
	cout << d[ed] << " " << c[ed];
	return 0;
 
}

Floyd算法

floyd的算法时间复杂度为O(n^3)，故顶点数n不能太大，一般不超过200

如果存在顶点k，使得以k作为中介点时顶点i和顶点j的当前最短距离缩短，则使用顶点k作为i和j的中介点。即当dis[i][k]+dis[k][j]

#include 
#include 
 
using namespace std;
 
const int INF = 100000000;
const int MAXV = 200;
int n, m;
int dis[MAXV][MAXV];
 
void Floyd() {
	for (int k = 0; k < n; k++) {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (dis[i][k] != INF && dis[k][j] != INF && dis[i][k] + dis[k][j] < dis[i][j]) {
					dis[i][j] = dis[i][k] + dis[k][j];
				}
			}
		}
	}
}
 
int main() {
	int u, v, w;
	fill(dis[0], dis[0] + MAXV * MAXV, INF);
	cin >> u >> v >> w;
	for (int i = 0; i < n; i++)dis[i][i] = 0;
	for (int i = 0; i < n; i++) {
		cin >> u >> v >> w;
		dis[u][v] = w;
	}
	Floyd();
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cout << dis[i][j] << " ";
		}
		cout << endl;
	}
 
	return 0;
}