2013
12-26

# Tourist

A lazy tourist wants to visit as many interesting locations in a city as possible without going one step further than necessary. Starting from his hotel, located in the north-west corner of city, he intends to take a walk to the south-east corner of the city and then walk back. When walking to the south-east corner, he will only walk east or south, and when walking back to the north-west corner, he will only walk north or west. After studying the city map he realizes that the task is not so simple because some areas are blocked. Therefore he has kindly asked you to write a program to solve his problem.
Given the city map (a 2D grid) where the interesting locations and blocked areas are marked, determine the maximum number of interesting locations he can visit. Locations visited twice are only counted once.

The first line in the input contains the number of test cases (at most 20). Then follow the cases. Each case starts with a line containing two integers, W and H (2 <= W, H <= 100), the width and the height of the city map. Then follow H lines, each containing a string with W characters with the following meaning:
‘.’ Walkable area
‘*’ Interesting location (also walkable area)
‘#’ Blocked area
You may assume that the upper-left corner (start and end point) and lower-right corner (turning point) are walkable, and that a walkable path of length H +W – 2 exists between them.

The first line in the input contains the number of test cases (at most 20). Then follow the cases. Each case starts with a line containing two integers, W and H (2 <= W, H <= 100), the width and the height of the city map. Then follow H lines, each containing a string with W characters with the following meaning:
‘.’ Walkable area
‘*’ Interesting location (also walkable area)
‘#’ Blocked area
You may assume that the upper-left corner (start and end point) and lower-right corner (turning point) are walkable, and that a walkable path of length H +W – 2 exists between them.

2
9 7
*........
.....**#.
..**...#*
..####*#.
.*.#*.*#.
...#**...
*........
5 5
.*.*.
*###.
*.*.*
.###*
.*.*.

7
8

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 105;
struct state {
int r, l, num;
}s[N * N];
int n, m, ans, begin, end, sum, f[N];

int getfather(int x) {
return x == f[x] ? x : f[x] = getfather(f[x]);
}

bool cmp(const state& a, const state& b) {
return a.num > b.num;
}

void init() {
for (int i = 0; i < m; i++)
scanf("%d%d%d", &s[i].l, &s[i].r, &s[i].num);
scanf("%d%d%d", &begin, &end, &sum);
for (int i = 1; i <= n; i++)
f[i] = i;
sort(s, s + m, cmp);
}

int kruskal() {
for (int i = 0; i < m; i++) {
int p = getfather(s[i].l), q = getfather(s[i].r);
if (p != q) {
f[p] = q;
if (getfather(begin) == getfather(end))
return s[i].num;
}
}
return 0;
}

int solve() {
if (sum % ans)
return sum / (ans - 1) + 1;
else
return sum / (ans - 1);
}

int main () {
int cas = 1;
while (scanf("%d%d", &n, &m), n || m ) {
init();
ans = kruskal();
printf("Scenario #%d\n", cas++);
printf("Minimum Number of Trips = %d\n\n", solve());
}
return 0;
}

1. “再把所有不和该节点相邻的节点着相同的颜色”，程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的