2018 BACS Regional Programming Contest F. Football Free Kick(map离散化)
Descripion
Solution
用map维护连续空区间端点。
可以用有三种状态的map维护，也可以用两个只有两种状态的map维护(很麻烦)
Code
map<int,bool>mpl, mpr;
int main(int argc, char const *argv[])
{int t;scanf ("%d",&t);int cas = 0;while (t--){mpl.clear();mpr.clear();int n,k,q;scanf("%d%d%d",&n,&k,&q);mpl[k+1] = true; mpr[n] = true;ll ans = 1;printf ("Case %d:\n",++cas);while(q--) {int x,y;scanf("%d %d",&x,&y);if(x==1) {if(!mpl[x+1]) {ans++;mpl[x] = mpr[x] = true; }else {mpl[x+1] = false;mpl[x] = true;}} else if(x==n) {if(!mpr[x-1]) {ans++;mpl[x] = mpr[x] = true; }else {mpr[x-1] = false;mpr[x] = true;}} else {if(mpl[x+1] && mpr[x-1]) {ans--;mpl[x+1] = false;mpr[x-1] = false;}else if(!mpl[x+1] && !mpr[x-1]) {ans++;mpl[x] = mpr[x] = true;}else {if(mpl[x+1] && !mpr[x-1]) {mpl[x+1] = false;mpl[x] = true;}else if(!mpl[x+1] && mpr[x-1]) {mpr[x-1] = false;mpr[x]=true;}}}if(y == 1) {if(mpl[y] && mpr[y]) {ans--;mpl[y] = mpr[y] = false;} else {mpl[y] = false; mpl[y+1] = true;}}else if(y == n) {if(mpl[y] && mpr[y]) {ans--;mpl[y] = mpr[y] = false;} else {mpr[y] = false; mpr[y-1] = true;}}else {if(mpl[y] && mpr[y]) {ans--;mpl[y] = mpr[y] = false;} else if(!mpl[y] && !mpr[y]) {ans++;mpr[y-1] = true;mpl[y+1] = true;} else {if(mpl[y]) mpl[y] = false, mpl[y+1] = true;else if(mpr[y]) mpr[y] = false, mpr[y-1] = true;}}printf("%lld\n", ans);}}return 0;
}
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