题目链接：点击打开链接
题意：
给定n个队伍的得分情况，输出随意一个可行解。
n个队伍随意2个队伍 a, b 间有且仅有一场比赛。
比赛结果分4种:
1、a +3， b +0
2、a +0， b +3
3、a +2， b +1
4、a +1， b +2
我们发现事实上每种结果2个队伍得分和总是3 且4种情况就是3的全部拆分和的形式。
所以我们把随意两个队伍组合成一个点。
把n个点连向源点，流上限为该队伍的得分。
对于1，2两个队伍
1 -> 点(1,2) 连流上限为3的边
2 -> 点(1,2) 连流上限为3的边
点(1,2) 到汇点连流上限为3的边。
若满流则有解。
1，2两个队伍间的情况就看点(1,2)的流入情况。
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
const int inf = (int)(1e9);
const int M = 200 + 2;
const int E = M * M * 6 + M * 3;
const int N = M * M + M + 5;
struct Edge {
int from, to, cap, nex;
}; int head[N], egnum;
Edge eg[E]; int a[M], n, st, ed;
int id[M][M], rid[N], tot;
int win[M][M]; void add(int u, int v, int cap, int rw = 0) {
Edge E = {u, v, cap, head[u]};
eg[egnum] = E;
head[u] = egnum++; Edge E2 = {v, u, rw, head[v]};
eg[egnum] = E2;
head[v] = egnum++;
}
int sign[N];
bool bfs(int from, int to) {
memset(sign, -1, sizeof sign);
sign[from] = 0; queue<int> q;
q.push(from);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = head[u]; i != -1; i = eg[i].nex) {
int v = eg[i].to;
if (sign[v] == -1 && eg[i].cap) {
sign[v] = sign[u] + 1;
q.push(v);
if (sign[to] != -1)
return true;
}
}
}
return false;
}
int Stack[N], top, cur[N];
int dicnic(int from, int to) {
int ans = 0;
while (bfs(from, to)) {
memcpy(cur, head, sizeof head);
int u = from;
top = 0;
while (true) {
if (u == to) {
int flow = inf, loc;
for (int i = 0; i < top; ++i)
if (flow > eg[Stack[i]].cap) {
flow = eg[Stack[i]].cap;
loc = i;
}
for (int i = 0; i < top; ++i) {
eg[Stack[i]].cap -= flow;
eg[Stack[i] ^ 1].cap += flow;
}
ans += flow;
top = loc;
u = eg[Stack[top]].from;
}
for (int i = cur[u]; i != -1; cur[u] = i = eg[i].nex)
if (eg[i].cap && (sign[u] + 1 == sign[eg[i].to]))
break;
if (cur[u] != -1) {
Stack[top++] = cur[u];
u = eg[cur[u]].to;
} else {
if (top == 0)
break;
sign[u] = -1;
u = eg[Stack[--top]].from;
}
}
}
return ans;
}
void init() {
memset(head, -1, sizeof head);
egnum = 0;
}
void pu(int x) {
if (x == 0) {
putchar('<');
} else if (x == 1) {
putchar('>');
} else if (x == 2) {
putchar('>');
putchar('=');
} else {
putchar('<');
putchar('=');
}
}
void work() {
int sum = 0, u, v;
for (int i = 0; i < n; ++i) {
scanf("%d", &a[i]);
sum += a[i];
}
tot = n;
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j) {
id[i][j] = ++tot;
rid[tot] = j;
}
st = ++tot;
ed = ++tot;
init();
for (int i = 0; i < n; ++i)
add(st, i, a[i]);
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j) {
add(i, id[i][j], 3);
add(j, id[i][j], 3);
add(id[i][j], ed, 3);
}
int g = dicnic(st, ed);
if (g != sum) {
puts("INCORRECT");
} else {
memset(win, -1, sizeof win);
puts("CORRECT");
for (int i = 0; i < egnum; ++i) {
if (eg[i].from < n && eg[i].to > n && eg[i].to < st) {
u = eg[i].from;
v = rid[eg[i].to];
if (u == v)
continue;
if (u > v)
std::swap(u, v);
if (win[u][v] == -1) {
if (eg[i].cap == 0) {
win[u][v] = 1; //1 = win, 0 = los, 2 = little win, 3 = little los
} else if (eg[i].cap == 1) {
win[u][v] = 2;
} else if (eg[i].cap == 2) {
win[u][v] = 3;
} else {
win[u][v] = 0;
}
}
}
}
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j) {
printf("%d ", i + 1);
pu(win[i][j]);
printf(" %d\n", j + 1);
}
}
}
int main() {
while (~scanf("%d", &n))
work();
return 0;
}