PTA 1003 Emergency

原题传送门》》

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:

Sample Output:

2 4
结尾无空行

思路：dijkstra算法，板子见文章末尾，这里不解释了。
但是需要额外对路径数量以及队伍数量进行计数。

关键是：当新收录了点u，如何操作才能记录路径数量？记录队伍数量？
做法是：逐个更新经过u的路径uv,查看新的dist[v]，假如dist[v]与之前的一样，则v的路径数量加上u的路径，队伍数量取uv的最大值
假如路径更短了，则路径数量和原u相同，队伍数量是u路径的队伍加上v城市的队伍

#include<cstdio>
#include<list>
using namespace std;int N,M,C1,C2;int L[500][500];
int dist[500];//到起点的距离
int troopNums[500];//某城市的队伍数量
int troopOnPath[500];//经过某路径的队伍数量
int pathNums[500];//到达该城市的路线数量
bool collected[500];//城市是否收入S int input(){for(int i = 0; i < 500; ++i){for(int j = 0; j < 500; ++j){L[i][j] = 0x3f3f3f3f;}}fill(dist,dist+500,0x3f3f3f3f);fill(collected,collected+500,false); scanf("%d%d%d%d",&N,&M,&C1,&C2);for(int i = 0; i < N; ++i){scanf("%d",&troopNums[i]);}for(int i = 0; i < M; ++i){int a, b, l;scanf("%d%d%d",&a,&b,&l);L[a][b] = l;L[b][a] = l;}dist[C1] = 0;troopOnPath[C1] = troopNums[C1];pathNums[C1] = 1; return 0;
}int dijk(){while(true){int u = -1;for(int i = 0; i < N; ++i){if(!collected[i]&&(u == -1 || dist[i] < dist[u])){u = i;}}if(u == -1) break;//找到未收录的距离最短的点，如果是-1，则说明没有找到这样的点，算法结束。collected[u] = true;for(int v = 0; v < N; ++v){if(!collected[v] && dist[u]+L[u][v] <= dist[v]){if(dist[u]+L[u][v] == dist[v]) {pathNums[v]+=pathNums[u];troopOnPath[v] = max(troopOnPath[v],troopOnPath[u]+troopNums[v]);}else{pathNums[v] = pathNums[u];troopOnPath[v] = troopOnPath[u]+troopNums[v];}dist[v] = dist[u]+L[u][v];}}}return 0;}int solve(){input();dijk();printf("%d %d\n",pathNums[C2],troopOnPath[C2]);return 0;
}
int main(){solve();return 0;
}

Dijkstra的板子：

const int MAX_V = 100;
const int INF = 0xffffff;
int cost[MAX_V][MAX_V];int d[MAX_V];//记录从出发点到该点的最短距离。bool used[MAX_V];//该点已经被收入S中int V;//顶点数void dijkstra(int s){fill(d,d+V,INF);
fill(used,used+V,false);
d[s] = 0;while(true){int v = -1;
for(int u = 0; u < V; ++u){if(!used[u] && (v == -1||d[u] < d[v]))
v = u;
}//从尚未使用过的点中找到一个距离最小的点 if(v == -1) break;used[v] = true;for(int u = 0; u < V; ++u){d[u] = min(d[u],d[v]+cost[v][u]);//更新v相连的点的值
}
}
}