1003. Emergency

1003. Emergency (25)

时间限制400 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueAs 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 markedon 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 inand 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 connectedby 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 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

题目的意思就是一张图，各边都有权值而且个点也有权值，求出给定两点之间最短路的条数并输出此路径上的点权值和的最大值。就是求出最短路的条数，若有多条相同的最短路输出最大的那个点权值和。很简单的题目，一个dfs就可以很好的解决掉。当然这个题目还可以继续增加点复杂度，就是把各条最短路的路径打印出来，这个也是很好实现的，加个pre数组，然后可以再来个dfs，当然也可以先来遍迪杰斯特拉。下面我们采用这两种方法来做，代码一：迪杰斯特克拉，代码而：dfs，当然用bfs也是可以做的，你也可以自己试一试。

代码一：迪杰斯特拉

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

int path[502][502];
int d[502];
int exist[502];
int rescue[502];
int cnt[502], r[502];

int n, m, s, e;

int max(int a, int b)
{
return a > b ? a : b;
}

int nearest()
{
int max = INT_MAX;
int index;

for(int i = 0; i < n; i ++)
{
if(!exist[i] && d[i] < max)
{
max = d[i];
index = i;
}
}
return index;
}

void shortest()
{
exist[s] = 1;

for(int i = 0; i < n; i ++)
{
d[i] = path[s][i];
cnt[i] = (d[i] == INT_MAX) ? 0 : 1;
r[i] = (d[i] == INT_MAX) ? INT_MIN : rescue[i] + rescue[s];
}

r[s] = rescue[s];
d[s] = 0;
cnt[s] = 1;

for(int i = 0; i < n-1; i ++)
{
int index = nearest();
exist[index] = 1;

for(int j = 0; j < n; j ++)
{
if(exist[j] || path[index][j] == INT_MAX)
continue;

if(d[index] + path[index][j] < d[j])
{
d[j] = d[index] + path[index][j];
cnt[j] = cnt[index];
r[j] =r[index] + rescue[j];
}
else if(d[index] + path[index][j] == d[j])
{
cnt[j] += cnt[index];
r[j] = max(r[index]+rescue[j], r[j]);
}
}
}
printf("%d %d\n", cnt[e], r[e]);
}

int main()
{
freopen("E://input.txt", "r", stdin);

scanf("%d%d%d%d", &n, &m, &s, &e);

for(int i = 0; i < n; i ++)
{
for(int j = 0; j < n; j ++)
path[i][j] = INT_MAX;
}

for(int i = 0; i < n; i ++)
scanf("%d", &rescue[i]);

for(int i = 0; i < m; i ++)
{
int a, b;
scanf("%d%d", &a, &b);
scanf("%d", &path[a][b]);
path[b][a] = path[a][b];
}

shortest();
return 0;
}

代码二：bfs

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

int rescue[502], exist[502], map[502][502];

int mind, cnt, maxt, n;//cnt记录有多少条最短路径，maxt记录多条相同的最短路中最大的那个各城市的救援队数量

void init(int n)
{
for(int i = 0; i < n; i ++)
{
exist[i] = 0;
for(int j = 0; j < n; j ++)
map[i][j] = INT_MAX;
}
}

void dfs(int p, int end, int dist, int numt)
{
if(p == end)
{
if(dist < mind)
{
cnt = 1;
mind = dist;
maxt = numt;
}
else if(dist == mind)
{
cnt ++;
if(maxt < numt)
maxt = numt;
}

return ;
}

if(dist > mind)
return ;//这个地方不剪枝的话最后一个case过不去的

for(int i = 0; i < n; i ++)
{
if(exist[i] == 0 && map[p][i] != INT_MAX)
{
exist[i] = 1;
dfs(i, end, dist+map[p][i], numt+rescue[i]);
exist[i] = 0;
}
}
}

int main()
{
freopen("E://input.txt", "r", stdin);
int m, st, end, x, y, d;
mind = INT_MAX;
cnt = 0;

scanf("%d%d%d%d", &n, &m, &st, &end);
init(n);
for(int i = 0; i < n; i ++)
scanf("%d", &rescue[i]);

while(m --)
{
scanf("%d%d%d", &x, &y, &d);
if(map[x][y] > d)
map[x][y] = map[y][x] = d;
}

dfs(st, end, 0, rescue[st]);
printf("%d %d\n", cnt, maxt);

return 0;
}