Fixing the Great Wall
https://odzkskevi.qnssl.com/b7d37f69f479d57e44735fc5d6403983?v=1508326275
【题解】
按照x排序
dp[i][j][0/1]表示从i到j全修好,当前在i/j的最小代价和已知的未修好的时间附加代价(td)转移即可，即
dp[i][j][0/1] - > d[i-1][j][0]
dp[i][j][0/1] - > d[i][j + 1][1]
为什么这样？想想不这样的情况，会发现都不如这样优秀
还有一个小处理，把起点虚成一个节点做
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #define min(a, b) ((a) < (b) ? (a) : (b)) 8 #define max(a, b) ((a) > (b) ? (a) : (b)) 9 10 inline void swap(int &a, int &b)11 {12 long long tmp = a;a = b;b = tmp;13 }14 15 inline void read(int &x)16 {17 x = 0;char ch = getchar(), c = ch;18 while(ch < '0' || ch > '9')c = ch, ch = getchar();19 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();20 }21 22 const int INF = 0x3f3f3f3f;23 const int MAXN = 10000 + 10;24 25 int n, v, s, sum[MAXN];26 struct Node27 {28 int x,c,d;29 }node[MAXN];30 31 int cmp(Node a, Node b)32 {33 return a.x < b.x;34 }35 36 //dp[i][j][0/1]表示修复i~j这段区间，当前在最左端（0）， 最右端（1）的37 //最小花费 38 39 double dp[MAXN][MAXN][2];40 41 int main()42 {43 while(scanf("%d %d %d", &n, &v, &s) != EOF && n && v && s)44 {45 for(register int i = 1;i <= n;++ i)46 read(node[i].x), read(node[i].c), read(node[i].d);47 node[n + 1].x = s;48 node[n + 1].c = 0;49 node[n + 1].d = 0;50 ++ n;51 std::sort(node + 1, node + 1 + n, cmp);52 for(register int i = 1;i <= n;++ i)53 sum[i] = sum[i - 1] + node[i].d;54 for(register int i = 1;i <= n;++ i)55 for(register int j = 1;j <= n;++ j)56 {57 if(i == j && node[i].x == s && node[i].c == 0 && node[i].d == 0) dp[i][j][0] = dp[i][j][1] = 0;58 else dp[i][j][0] = dp[i][j][1] = INF;59 }60 for(register int len = 1;len <= n;++ len)61 {62 for(register int i = 1;i <= n;++ i)63 {64 int j = i + len - 1;65 dp[i - 1][j][0] = min(dp[i - 1][j][0],66 min(dp[i][j][0] + (double)(node[i].x - node[i - 1].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[i - 1].c,67 dp[i][j][1] + (double)(node[j].x - node[i - 1].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[i - 1].c));68 dp[i][j + 1][1] = min(dp[i][j + 1][1], 69 min(dp[i][j][0] + (double)(node[j + 1].x - node[i].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[j + 1].c,70 dp[i][j][1] + (double)(node[j + 1].x - node[j].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[j + 1].c));71 }72 }73 printf("%d\n", min((int)dp[1][n][0], (int)dp[1][n][1]));74 }75 return 0;76 }