题目链接:点击打开链接
题意很容易理解,基本上这个题意就是让你去考虑无向图的最小割,无奈只会用模版,还没想好怎么用堆或者斐波那契堆优化Stoer-Wagner算法,TLE......而且百度把这题的时间放宽到了2000ms,这样还超时,不会优化真尴尬......SW算法用堆优化一般适用于稀疏图,如果是稠密图,优化还可能更费时,看此题的数据,点的个数取值为3 * 10^3,边的个数为10^5,应该是稀疏图的测试超了时。
而用了并查集 + 记录点的权和,用并查集判断是否是连通图,如果不连通,直接输出0,如果连通,输出最小权和,即把权和最小的点分割出来,竟然对了......一开始也想了并查集,写了一点感觉不靠谱才换的SW,没想到此题这么坑。不过有时间还是得学会优化SW算法。
// 度度熊的王国战略2.cpp 运行/限制:140ms/2000ms
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int n, m, fa[3005], sum[3005];
int get(int x) {
return fa[x] = fa[x] == x ? x : get(fa[x]);
}
void merge(int x, int y) {
int a = get(x);
int b = get(y);
if (a != b) {
fa[b] = a;
}
}
void init() {
for (int i = 1; i <= n; i++) {
fa[i] = i;
}
}
int main(){
int a, b, c, cnt;
while (scanf("%d%d", &n, &m) != EOF) {
init();
cnt = n - 1;
memset(sum, 0, sizeof(sum));
for (int i = 0; i < m; i++) {
scanf("%d%d%d", &a, &b, &c);
if (a == b) continue;
sum[a] += c;
sum[b] += c;
int x = get(a);
int y = get(b);
if (x != y) {
merge(x, y);
cnt--;
}
}
if (cnt) {
printf("0\n");
}
else{
sort(sum + 1, sum + n + 1);
printf("%d\n", sum[1]);
}
}
return 0;
}
// 度度熊的王国战略.cpp Time Limit Exceeded
#include "stdafx.h"
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define INF 0X3f3f3f3f
int vis[3005];
int wet[3005];
int combine[3005];
int map[3005][3005];
int S, T, minCut, N, M;
void Search() {
int i, j, Max, tmp;
memset(vis, 0, sizeof(vis));
memset(wet, 0, sizeof(wet));
S = T = -1;
for (i = 1; i <= N; i++) {
Max = -INF;
for (j = 1; j <= N; j++) {
if (!combine[j] && !vis[j] && wet[j] > Max) {
tmp = j;
Max = wet[j];
}
}
if (T == tmp) return;
S = T; T = tmp;
minCut = Max;
vis[tmp] = 1;
for (j = 1; j <= N; j++) {
if (!combine[j] && !vis[j]) {
wet[j] += map[tmp][j];
}
}
}
}
int Stoer_Wagner() {
int i, j;
memset(combine, 0, sizeof(combine));
int ans = INF;
for (i = 1; i < N; i++) {
Search();
if (minCut < ans) ans = minCut;
if (ans == 0) return 0;
combine[T] = 1;
for (j = 1; j <= N; j++) {
if (!combine[j]) {
map[S][j] += map[T][j];
map[j][S] += map[j][T];
}
}
}
return ans;
}
int main() {
int a, b, c;
while (scanf("%d%d", &N, &M) != EOF) {
memset(map, 0, sizeof(map));
while (M--) {
scanf("%d%d%d", &a, &b, &c);
map[a][b] += c;
map[b][a] += c;
}
printf("%d\n", Stoer_Wagner());
}
return 0;
}