USACO: Sweet Butter

Sweet Butter

Greg Galperin -- 2001

Farmer John has discovered the secret to making the sweetest butter in all of Wisconsin: sugar. By placing a sugar cube out in the pastures, he knows the N (1 <= N <= 500) cows will lick it and thus will produce super-sweet butter which can be marketed at better prices. Of course, he spends the extra money on luxuries for the cows.

FJ is a sly farmer. Like Pavlov of old, he knows he can train the cows to go to a certain pasture when they hear a bell. He intends to put the sugar there and then ring the bell in the middle of the afternoon so that the evening's milking produces perfect milk.

FJ knows each cow spends her time in a given pasture (not necessarily alone). Given the pasture location of the cows and a description of the paths the connect the pastures, find the pasture in which to place the sugar cube so that the total distance walked by the cows when FJ rings the bell is minimized. FJ knows the fields are connected well enough that some solution is always possible.

PROGRAM NAME: butter

INPUT FORMAT

• Line 1: Three space-separated integers: N, the number of pastures: P (2 <= P <= 800), and the number of connecting paths: C (1 <= C <= 1,450). Cows are uniquely numbered 1..N. Pastures are uniquely numbered 1..P.
• Lines 2..N+1: Each line contains a single integer that is the pasture number in which a cow is grazing. Cow i's pasture is listed on line i+1.
• Lines N+2..N+C+1: Each line contains three space-separated integers that describe a single path that connects a pair of pastures and its length. Paths may be traversed in either direction. No pair of pastures is directly connected by more than one path. The first two integers are in the range 1..P; the third integer is in the range (1..225).

SAMPLE INPUT (file butter.in)

3 4 5
2
3
4
1 2 1
1 3 5
2 3 7
2 4 3
3 4 5

INPUT DETAILS

This diagram shows the connections geometrically:

          P2  
 P1 @--1--@ C1
     /    |/
      /   | /
       5  7  3
        / |   /
         /|    / C3
       C2 @--5--@
          P3    P4

OUTPUT FORMAT

• Line 1: A single integer that is the minimum distance the cows must walk to a pasture with a sugar cube.

SAMPLE OUTPUT (file butter.out)

8

OUTPUT DETAILS:

Putting the cube in pasture 4 means: cow 1 walks 3 units; cow 2 walks 5
units; cow 3 walks 0 units -- a total of 8.

 
思路：

试过floyd和dijkstra，结果都是TLE，于是直接在dijkstra的基础上加了个堆优化，不过写得太“直接”，效率还是很低，0.954s最后两组

/* ID: geterns1 PROG: butter LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; const int infi = 1000000; const int size = 800 + 1; //原始数据 int edge[size][size], realSize; //目标数组 const int tarSize = 500; int maps[tarSize]; int dist[tarSize][size]; int path[tarSize][size]; //辅助堆 int heap[size], curSize; inline void qSwap(int &a, int &b){ a ^= b; b ^= a; a ^= b; } inline int fath(int idx){ return (idx - 1) / 2; } inline int left(int idx){ return 2 * idx + 1; } inline int rght(int idx){ return 2 * (idx + 1); } void siftDw(int tar, int idx){ int tarIdx = idx; while (true){ if (left(idx) < curSize && dist[tar][heap[left(idx)]] < dist[tar][heap[tarIdx]]) tarIdx = left(idx); if (rght(idx) < curSize && dist[tar][heap[rght(idx)]] < dist[tar][heap[tarIdx]]) tarIdx = rght(idx); if (tarIdx ^ idx){ qSwap(heap[idx], heap[tarIdx]); idx = tarIdx; } else return; } } void siftUp(int tar, int idx){ while (idx){ if (dist[tar][heap[fath(idx)]] > dist[tar][heap[idx]]){ qSwap(heap[fath(idx)], heap[idx]); idx = fath(idx); } else return; } } //初始化图 void initail(){ for (int i = 1; i <= realSize; i ++) for (int j = 1; j <= realSize; j ++) edge[i][j] = (i == j ? 0 : infi); } //迪克斯特拉算法求最短路（无向图要求矩阵对称） void dijkstra(int tar){ //初始化辅助数组和目标数组 curSize = 0; for (int i = 1; i <= realSize; i ++){ dist[tar][i] = edge[maps[tar]][i]; if (i != maps[tar]){ heap[curSize ++] = i; siftUp(tar, curSize - 1); } } //确定n-1条路径 for (int i = 0; i < realSize - 1; i ++){ //弹出下一最短路 int u = heap[0]; heap[0] = heap[-- curSize]; siftDw(tar, 0); //更新和u指向的顶点 for (int j = 0; j < curSize; j ++){ if (edge[u][heap[j]] < infi && dist[tar][u] + edge[u][heap[j]] < dist[tar][heap[j]]){ dist[tar][heap[j]] = dist[tar][u] + edge[u][heap[j]]; siftUp(tar, j); } } } } int main(){ freopen("butter.in", "r", stdin); freopen("butter.out", "w", stdout); int n, c, x, y; scanf("%d %d %d/n", &n, &realSize, &c); initail(); for (int i = 0; i < n; i ++) scanf("%d/n", maps + i); while (c --){ scanf("%d %d ", &x, &y); scanf("%d/n", &edge[x][y]); edge[y][x] = edge[x][y]; } for (int i = 0; i < n; i ++) dijkstra(i); // int min = infi, sum; for (int i = 1; i <= realSize; i ++){ sum = 0; for (int j = 0; j < n; j ++) sum += dist[j][i]; if (sum < min) min = sum; } printf("%d/n", min); fclose(stdin); fclose(stdout); return 0; }