POJ 1797 Heavy Transportation
題目大意:
存在圖中1到n號點的路徑,求多條路徑中最小權值最大的一條,輸出最大的最小權值
具體思路:
改寫最短路解法或者最大生成樹(把權值變爲負數,直接用最小生成樹算法,結果取絕對值)
詳細思路可見這篇博客POJ2253,它是求最小的最大權值。
下方代碼爲改寫最短路
具體代碼:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int N = 1005;
const int INF = 1e5;
int maps[N][N];
int visit[N];
int d[N]; //終點爲i的最大最小權值
int cross, street;
void init()
{
for (int i = 1; i <= cross; i++)
{
visit[i] = 0;
}
for (int i = 1; i <= cross; i++)
for (int j = 1; j <= cross; j++)
maps[i][j] = -1; //這裏要改寫
}
void dijkstra()
{
for (int i = 1; i <= cross; i++)
{
d[i] = maps[1][i];
}
d[1] = 0;
visit[1] = 1;
for (int i = 1; i <= cross; i++)
{
int t = -1;
for (int j = 2; j <= cross; j++)
{
if (!visit[j] && (t == -1 || d[t] < d[j])) //這裏要改寫
t = j;
}
if (t == -1)return;
visit[t] = 1;
for (int j = 2; j <= cross; j++)
{
if (!visit[j] && d[j] < min(d[t], maps[t][j])) //這裏要改寫
d[j] = min(d[t], maps[t][j]);
}
}
}
int main()
{
int n;
int cnt = 1;
cin >> n;
for (int i = 1; i <= n; i++)
{
int x, y, w;
cin >> cross >> street;
init();
for (int j = 1; j <= street; j++)
{
cin >> x >> y >> w;
if (w > maps[x][y]) //考慮重邊,這裏要改寫
maps[x][y] = maps[y][x] = w;
}
dijkstra();
printf("Scenario #%d:\n", cnt);
printf("%d\n\n", d[cross]);
cnt++;
}
return 0;
}