Zeray Rice

..

[USACO 3.2] Sweet Butter

解析

题目意思很清楚… 算以每个点为终点, 其他点到这个点的最短路之和..的最小值是多少…

求最短路用的SPFA.. 嗯.. Floyd 时间复杂度太高… 不适合.. 不过我只是扔了一个裸的SPFA 上去..也没见有多慢啊… 难道是USACO的评测机变好了?

坑爹的一点是 这道题最后调了半天.. USACO 评测的时候和我本地的输出不一样.. 查了半天最后发现 spfa 函数里面的 atqueue 没有 memset 成 flase … 导致蛋疼..

经 vczh 巨巨指点… 原来如果在建立数组的时候 bool array[80] = {0} 就行了.. 之前一直以为这个操作是等同于 array[0] = 0 的… 巨巨解释说… 这个是 C++ 的黑细节… 如果是 int array[80] = {1} … 结果就是 array[0] = 1 了… 只有 {0} 是例外的…

代码

USACO 3.2 Sweet Butter raw
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/*
USER: fanzeyi1
LANG: C++
TASK: butter
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#define FARM_NUM 802

using namespace std;

int cow, farm, road;
int map[FARM_NUM][FARM_NUM];
int cost[FARM_NUM][FARM_NUM];
int farms[FARM_NUM];

int result = 0x7FFFFFFF;

int spfa(int begin) {
    queue<int> q;
    bool atqueue[farm+1];
    int dist[farm+1];
    int ans = 0;
    int now;
    memset(dist, 0x7F, sizeof(dist));
    memset(atqueue, 0, sizeof(atqueue));
    dist[begin] = 0;
    atqueue[begin] = true;
    q.push(begin);
    while(!q.empty()) {
        now = q.front();
        q.pop();
        atqueue[now] = false;
        for(int i = 1; i <= map[now][0]; i++) {
            if(dist[now] + cost[now][map[now][i]] < dist[map[now][i]]) {
                dist[map[now][i]] = dist[now] + cost[now][map[now][i]];
                if(!atqueue[map[now][i]]) {
                    q.push(map[now][i]);
                    atqueue[map[now][i]] = true;
                }
            }
        }
    }
    for(int i = 1; i <= farm; i++)  {
        ans = ans + dist[i] * farms[i];
    }
    printf("\n");
    return ans;
}

int main() {
    FILE *fin = fopen("butter.in", "r");
    FILE *fout = fopen("butter.out", "w");
    int a, b, c;
    memset(farms, 0, sizeof(farms));
    fscanf(fin, "%d %d %d", &cow, &farm, &road);
    for(int i = 0; i < cow; i++) {
        fscanf(fin, "%d", &a);
        farms[a] = farms[a] + 1;
    }
    for(int i = 0; i < road; i++) {
        fscanf(fin, "%d %d %d", &a, &b, &c);
        map[a][++map[a][0]] = b; cost[a][b] = c;
        map[b][++map[b][0]] = a; cost[b][a] = c;
    }
    for(int i = 1; i <= farm; i++) {
        a = spfa(i);
        if(a < result) {
            result = a;
        }
    }
    fprintf(fout, "%d\n", result);
    return 0;
}
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Executing...
   Test 1: TEST OK [0.000 secs, 8216 KB]
   Test 2: TEST OK [0.000 secs, 8216 KB]
   Test 3: TEST OK [0.000 secs, 8216 KB]
   Test 4: TEST OK [0.011 secs, 8216 KB]
   Test 5: TEST OK [0.011 secs, 8216 KB]
   Test 6: TEST OK [0.032 secs, 8216 KB]
   Test 7: TEST OK [0.086 secs, 8216 KB]
   Test 8: TEST OK [0.140 secs, 8216 KB]
   Test 9: TEST OK [0.194 secs, 8216 KB]
   Test 10: TEST OK [0.194 secs, 8216 KB]

All tests OK.

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