Codeforces 1131D Gourmet choice——差分约束

题意：给定一张比较表，给行列赋值使得比较表成立，并且最大值最小

思路：差分约束，求解最长路

/**
* x > y : x >= y + 1;
* x < y : x <= y - 1;
* x = y : x >= y x <= y
* maxlongpath: u -> v :
* if (dis[v] < dis[u] + cost) dis[v] = dis[u] + cost :
* dis[v] >= dis[u] + cost(u, v);
* x > y : y -> x cost = 1;
* x < y : x -> y cost = 1;
* x = y : x -> y, cost = 0, y -> x, cost = 0;
*/
#include 
using namespace std;
const int maxn = 1005;
int n, m;
char c;
typedef pair P;
vector
 vec[maxn*2];
int du[maxn*2];
int dis[maxn*2], vis[maxn*2], cnt[maxn*2];
const int INF = 0x3f3f3f3f;
bool spfa(int s) {
    memset(dis, -1, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    queue que;
    que.push(s);
    dis[s] = 0;
    vis[s] = 1;
    cnt[s] = 1;
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        vis[u] = 0;
        for (int i = 0; i < vec[u].size(); i++) {
            int v = vec[u][i].first, cost = vec[u][i].second;
            if (dis[v] < dis[u] + cost) {
                dis[v] = dis[u] + cost;
                if (!vis[v]) {
                    cnt[v]++;
                    if (cnt[v] >= n+m+1) return 0;
                    vis[v] = 1;
                    que.push(v);
                }
                else {

                }
            }
        }
    }
    return 1;
}
int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) {
        getchar();
        for (int j = 1; j <= m; j++) {
            scanf("%c", &c);
            if (c == '>') {
                vec[j+n].push_back(make_pair(i, 1));
            }
            else if (c == '<') {
                vec[i].push_back(make_pair(j+n, 1));
            }
            else {
                vec[j+n].push_back(make_pair(i, 0));
                vec[i].push_back(make_pair(j+n, 0));
            }
        }
    }
    for (int i = 1; i <= n + m; i++) {
        vec[0].push_back(make_pair(i, 1));
    }
    bool ok = spfa(0);
    if (ok) {
        for (int i = 1; i <= n + m; i++) {
            if (dis[i] == -1) {
                ok = false;
                break;
            }
        }
    }
    if (ok) {
        puts("Yes");
        for (int i = 1; i <= n; i++) {
            printf("%d ", dis[i]);
        }
        printf("\n");
        for (int i = n + 1; i <= n + m; i++) {
            printf("%d ", dis[i]);
        }
        printf("\n");
    }
    else {
        puts("No");
    }
    return 0;
}