POJ 1733 Parity game——并查集 + 离散化

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#include <cstdio>#include <algorithm>using namespace std;const int maxn = 1e4 + 10;int T, N, M, par[maxn], val[maxn], a[maxn];struct Date {    char str[10];    int x, y;}date[maxn];int Query(int x) {    if (x != par[x]) {        int temp = par[x];        par[x] = Query(par[x]);        val[x] = val[x] ^ val[temp];    }    return par[x];}int main() {    while (~scanf("%d %d", &N, &M)) {        int cnt = 0;        for (int i = 1; i <= M; i++) {            scanf("%d%d%s", &date[i].x, &date[i].y, date[i].str); date[i].x--;            a[++cnt] = date[i].x; a[++cnt] = date[i].y;        }        sort(a + 1, a + 1 + cnt);        cnt = unique(a + 1, a + 1 + cnt) - a;        for (int i = 1; i <= cnt; i++) par[i] = i, val[i] = 0;        int ans = 0;        for (int i = 1; i <= M; i++) {            int x = lower_bound(a + 1, a + 1 + cnt, date[i].x) - a;            int y = lower_bound(a + 1, a + 1 + cnt, date[i].y) - a;            int a = Query(x), b = Query(y);            if (a == b) {                if ((val[x] ^ val[y]) % 2 == 0 && date[i].str[0] == 'o') break;                if ((val[x] ^ val[y]) % 2 == 1 && date[i].str[0] == 'e') break;                ans++;            }            else {                if (date[i].str[0] == 'o') {                    par[a] = b;                    val[a] = val[x] ^ val[y] ^ 1;                }                else {                    par[a] = b;                    val[a] = val[x] ^ val[y];                }                ans++;            }        }        printf("%d\n", ans);    }}
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