题意:有n只队伍(n是2的整数幂)打淘汰赛,每一轮都是两两配对,胜利者进入下一轮。每一只队伍的实力都是确定的,你最喜欢编号为一的队伍,设计一个方案让队伍一一定胜利。
思路:这个题目有两个关键性的条件
条件1:1号队伍至少能打败一半的队伍。
条件2:对于不能直接打败的队伍t,存在能够打败的队伍t1,t1能够打败t。
这两个条件是至关重要的,设计三个阶段
阶段1:尽可能的消灭队伍1无法打败的队伍
阶段2:在还没匹配的队伍中找到一个1号队伍能够打败的
阶段3:将剩余的队伍进行匹配
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#define MAXN 1100
#define MAXE 210
#define INF 10000000
#define MOD 1000000007
#define LL long long
#define pi acos(-1.0)
using namespace std;
char str[MAXN][MAXN];
vector<int> win;
vector<int> lose;
int main() {
std::ios::sync_with_stdio(false);
int n;
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; ++i) {
scanf("%s", str[i] + 1);
}
win.clear();
lose.clear();
for (int i = 2; i <= n; ++i) {
if (str[1][i] == '1') {
win.push_back(i);
} else {
lose.push_back(i);
}
}
int m = n;
while (m >>= 1) {
vector<int> win2, lose2, final2;
for (int i = 0; i < lose.size(); ++i) {
bool flag = false;
for (int j = 0; j < win.size(); ++j) {
if (win[j] != -1 && str[win[j]][lose[i]] == '1') {
printf("%d %d\n", win[j], lose[i]);
win2.push_back(win[j]);
win[j] = -1;
flag = true;
break;
}
}
if (!flag) {
final2.push_back(lose[i]);
}
}
bool flag = false;
for (int i = 0; i < win.size(); ++i) {
if (win[i] != -1 && !flag) {
flag = true;
printf("1 %d\n", win[i]);
win[i] = -1;
} else if (win[i] != -1 && flag) {
final2.push_back(win[i]);
}
}
for (int i = 0; i < final2.size(); i += 2) {
printf("%d %d\n", final2[i], final2[i + 1]);
if (str[final2[i]][final2[i + 1]] == '1') {
if (str[1][final2[i]] == '1') {
win2.push_back(final2[i]);
} else {
lose2.push_back(final2[i]);
}
} else {
if (str[1][final2[i + 1]] == '1') {
win2.push_back(final2[i + 1]);
} else {
lose2.push_back(final2[i + 1]);
}
}
}
win = win2;
lose = lose2;
}
}
return 0;
}
/*
4
0110
0011
0000
1010
8
00111010
10101111
00010010
01000101
00110010
10101011
00010000
10101010
*/