OUC _ preparations for the provincial competition _ #5 _ F question poj3662

Http://poj.org/problem? Id = 3662 question: Find a path from 1 to n. You can select k paths for free and ask what is the maximum edge length after k entries are exceeded in the path. During the competition, I came up with a correct algorithm: Dual answer + DP determination, but I fell down one place, and the dual answer was reversed several times, but it didn't happen again, when we get back, we can change the binary mode. dp [I] [j] indicates that the maximum value of j paths greater than L is obtained when we reach the I point. It is also possible to directly solve the problem of no solution and 0 through dp, directly expressing the status as dp [I] [j] indicates that the maximum code1 value reaches the I point using j free routes, or two people wrote a part of [code lang = "cpp"] bool spfa (int L) {memset (vv, 0, sizeof (vv); memset (dp,-1, sizeof (dp); while (! Q. empty () q. pop (); bool ffff; ffff = false; node u, tu; u. v = s; u. big = 0; vv [s] [0] = 1; dp [s] [0] = 0; q. push (u); while (! Q. empty () {u = q. front (); q. pop (); if (u. v = n) {if (u. big = k) {if (dp [n] [k]! =-1) {ans = min (ans, dp [n] [k]) ;}} if (u. big & gt; = k) ffff = true; vv [u. v] [u. big] = 0; continue;} for (int I = head [u. v]; I! =-1; I = edge [I]. next) {int v = edge [I]. v; int tmp; if (edge [I]. val & lt; = L) tmp = max (dp [u. v] [u. big], edge [I]. val); else tmp = dp [u. v] [u. big]; if (edge [I]. val & gt; L) {if (tmp & lt; dp [v] [u. big + 1] | dp [v] [u. big + 1] =-1) & amp; (u. big + 1 & lt; = k) {dp [v] [u. big + 1] = tmp; if (! Vv [v] [u. big + 1]) {tu. v = v; tu. big = u. big + 1; q. push (tu); vv [v] [u. big + 1] = 1 ;}}} else {if (tmp & lt; dp [v] [u. big] | dp [v] [u. big] =-1) {dp [v] [u. big] = tmp; if (! Vv [v] [u. big]) {tu. v = v; tu. big = u. big; q. push (tu); vv [v] [u. big] = 1 ;}}} vv [u. v] [u. big] = 0;} return ffff;} bool vis [maxn]; int que [maxn]; int dis [maxn]; bool check () {memset (vis, false, sizeof (vis); memset (dis,-1, sizeof (dis); int hh = 0, tail = 1; que [1] = 1; vis [1] = 1; dis [1] = 0; while (hh & lt; tail) {int u = que [++ hh]; for (int j = head [u]; j! =-1; j = edge [j]. next) {int v = edge [j]. v; if (dis [u] + 1 & lt; dis [v] | dis [v] =-1) {dis [v] = dis [u] + 1; if (! Vis [v]) {que [++ tail] = v; vis [v] = 1 ;}} vis [u] = 0 ;}if (dis [n]! =-1 & amp; dis [n] & lt; = k) return true; return false;} int main () {int u, v, val; while (scanf (& quot; % d & quot;, & amp; n, & amp; p, & amp; k )! = EOF) {cnt = 0; s = 1; memset (head,-1, sizeof (head); int mmjh = 0; for (int I = 0; I & lt; p; I ++) {scanf (& quot; % d & quot;, & amp; u, & amp; v, & amp; val); addedge (u, v, val); mm1_= max (mm1_, val);} if (check () {printf (& quot; 0 \ n & quot;); continue;} int left, right, mid; left = 1; right = mmjh; ans = inf; while (left & lt; right) {mid = (left + right)/2; bool flag; flag = spfa (mid); if (flag = 0 ){ Left = mid + 1;} else {right = mid;} if (ans = inf) {ans =-1;} printf (& quot; % d \ n & quot;, ans);} return 0;} [/code] code2 [code lang = "cpp"] void spfa () {memset (vv, 0, sizeof (vv); for (int I = 0; I & lt; = n; I ++) {for (int j = 0; j & lt; = k; j ++) {dp [I] [j] = inf ;}} while (! Q. empty () q. pop (); node u, tu; u. v = s; u. big = 0; vv [s] [0] = 1; dp [s] [0] = 0; q. push (u); while (! Q. empty () {u = q. front (); q. pop (); if (u. v = n) {// printf (& quot; % d \ n & quot;, u. v, u. big, dp [u. v] [u. big]); ans = min (ans, dp [u. v] [u. big]);} for (int I = head [u. v]; I! =-1; I = edge [I]. next) {int v = edge [I]. v; int tmp; tmp = max (dp [u. v] [u. big], edge [I]. val); if (dp [v] [u. big] & gt; tmp) {dp [v] [u. big] = tmp; if (! Vv [v] [u. big]) {tu. v = v; tu. big = u. big; q. push (tu); vv [v] [u. big] = 1 ;}} if (u. big & lt; k & amp; dp [u. v] [u. big] & lt; dp [v] [u. big + 1]) {dp [v] [u. big + 1] = dp [u. v] [u. big]; if (! Vv [v] [u. big + 1]) {tu. v = v; tu. big = u. big + 1; q. push (tu); vv [v] [u. big + 1] = 1 ;}}// printf (& quot; \ n & quot;); vv [u. v] [u. big] = 0 ;}} int main () {int u, v, val; while (scanf (& quot; % d & quot;, & amp; n, & amp; p, & amp; k )! = EOF) {cnt = 0; s = 1; memset (head,-1, sizeof (head); for (int I = 0; I & lt; p; I ++) {scanf (& quot; % d & quot;, & amp; u, & amp; v, & amp; val); addedge (u, v, val) ;}ans = inf; spfa (); if (ans = inf) {ans =-1;} printf (& quot; % d \ n & quot;, ans);} return 0;} [/code]